A body thrown upwards from a stand of height

voltaire101 · Nov 27, 2011

I want to check whether my solution is right or not... beside I have made some correction to the problem so that it will be consistent ( I want to check also this correction).

1. The problem statement, all variables and given/known data

A body is thrown vertically upwards from a stand of height 30 meters above the ground. If the height h(t) above the ground after t seconds is given by : h(t) =20t-5t^2 +25 meters . Find:

(a) The initial velocity and acceleration and also the velocity and acceleration after t seconds.

(b) The highest point reached by the body.

(c) When the body strikes the ground and its velocity then.

(d) Suppose that instead of being thrown from a stand the body was thrown vertically from the ground with the same acceleration and initial velocity . When will the body strike the ground and find its velocity then.

******************************
This problem is incorrectly formulated:- (I am not sure about that)

I have copied this problem as it is,, and I think that there is a mistake in the problem formulation

Since the stand's height is 30 meters above the ground, the equation should be formulated like this
h(t) =20t-5t^2 +30 meters.

OR instead (that what I will do) correct the height of the stand from 30 to 25 so that the corrected version will be:

A body is thrown vertically upwards from a stand of height 25 meters above the ground. If the height h(t) above the ground after t seconds is given by : h(t) =20t-5t^2 +25 meters.

******************************
2. Relevant equations

3. The attempt at a solution

(a) The initial velocity and acceleration and also the velocity and acceleration after t seconds.

the velocity:

v(t) = (dh/dt) = 20-10t m/s

initial velocity is when t=0

v(0) = 20 m/s

the acceleration:

a(t) = (dv/dt) = -10 m/s^2

so a=-10 m/s^2

....................................

(b) The highest point reached by the body.

the highest point is when v(t) = 20-20t = 0

20-10t=0

t = 2 seconds

the highest point can be given using h(t) =20t-5t^2 +25

h(2) =20(2)-5(2)^2 +25 = 45 m

......................................

(c) When the body strikes the ground and its velocity then.

the body strikes the ground when h(t) =20t-5t^2 +25 = 0

20t-5t^2 +25 = 0

using the quadratic formula

t = -1 seconds and t = 5 seconds

**************************
I want to know what each result represents, and what is the meaning of the minus sign in (t=-1)
**************************
the velocity can be given

v(t) = 20-10t m/s

v(-1) = 30 m/s

v(5) = -30 m/s

**********************
I want to know what each result represents, and what is the meaning of the minus sign in v(5) = -30 m/s
**********************
.............................

(d) Suppose that instead of being thrown from a stand the body was thrown vertically from the ground with the same acceleration and initial velocity . When will the body strike the ground and find its velocity then.

Here I changes the equation of height to be
h(t) =20t-5t^2

the equation of velocity (not changed)
v(t) = 20-10t m/s

the body will strike the ground when h(t) =20t-5t^2 = 0

t(20-5t) = 0

t=0 seconds and t = 4 seconds

**************************
I want to know what each result represents.
**************************

the velocity can be given v(t) = 20-10t m/s

v(0) = 20 m/s

v(4) = -20 m/s

**************************
I want to know what each result represents. and what is the meaning of the minus sign in v(4) = -20 m/s
**************************

danielakkerma · Nov 27, 2011

Hi there.
Unless there's some convoluted trick nestled in the query, such as having the object thrown with some initial displacement, while already on the stand, I think you're positively right in the corrections you provided, though +25 seems more accurate, since it produces very neat and tidy results(in the quadratic equation).
Next, it seems to me that you've solved everything accurately; Note an easy form of validation: your acceleration a=-10, hence -g, which what you would expect for a body under the 'auspices' of earth's gravity;
As for the two roots, try sketching the h(t) parabola, taking t, the horizontal axis, from some negative value to a positive one, and observing the result, Keep in mind, that when it comes to time, it is our physical convention that for a physical process, some event starts at 0. However, there "was"(perhaps) something before, and if we allow it to continue(and conditions permit), there will be some "after"(literally, after the object has hit the ground in this case).
The negative results when plugging in the velocity are evident, someone has already chosen the sign convention for you in this problem; That is, a-the acceleration is already taken to be negative, so when the object returns to the ground, it has a velocity matching the -y direction.
Hope that clears things up,
Daniel

PeterO · Nov 27, 2011

voltaire101 said: ↑

I want to check whether my solution is right or not... beside I have made some correction to the problem so that it will be consistent ( I want to check also this correction).

1. The problem statement, all variables and given/known data

A body is thrown vertically upwards from a stand of height 30 meters above the ground. If the height h(t) above the ground after t seconds is given by : h(t) =20t-5t^2 +25 meters . Find:

(a) The initial velocity and acceleration and also the velocity and acceleration after t seconds.

(b) The highest point reached by the body.

(c) When the body strikes the ground and its velocity then.

(d) Suppose that instead of being thrown from a stand the body was thrown vertically from the ground with the same acceleration and initial velocity . When will the body strike the ground and find its velocity then.

******************************
This problem is incorrectly formulated:- (I am not sure about that)

I have copied this problem as it is,, and I think that there is a mistake in the problem formulation

Since the stand's height is 30 meters above the ground, the equation should be formulated like this
h(t) =20t-5t^2 +30 meters.

OR instead (that what I will do) correct the height of the stand from 30 to 25 so that the corrected version will be:

A body is thrown vertically upwards from a stand of height 25 meters above the ground. If the height h(t) above the ground after t seconds is given by : h(t) =20t-5t^2 +25 meters.

******************************
2. Relevant equations

3. The attempt at a solution

(a) The initial velocity and acceleration and also the velocity and acceleration after t seconds.

the velocity:

v(t) = (dh/dt) = 20-10t m/s

initial velocity is when t=0

v(0) = 20 m/s

the acceleration:

a(t) = (dv/dt) = -10 m/s^2

so a=-10 m/s^2

....................................

(b) The highest point reached by the body.

the highest point is when v(t) = 20-20t = 0

20-10t=0

t = 2 seconds

the highest point can be given using h(t) =20t-5t^2 +25

h(2) =20(2)-5(2)^2 +25 = 45 m

......................................

(c) When the body strikes the ground and its velocity then.

the body strikes the ground when h(t) =20t-5t^2 +25 = 0

20t-5t^2 +25 = 0

using the quadratic formula

t = -1 seconds and t = 5 seconds

**************************
I want to know what each result represents, and what is the meaning of the minus sign in (t=-1)

Suppose there is a person on the ground behind the stand, and they project an object vertically at 30 m/s but 1 second before the one you are calculating for. It would travel exactly level with your projectile - from the time it passes the top of the stand until it hits the ground.
There is nothing sinister about the negative time - it comes about due to you defining time zero as the time when you threw your object.

You could have said that you threw your object at time 13 seconds - in which case it would land at time 18 seconds, and this parallel object would have to have been thrown from ground level at time 12 seconds.

**************************
the velocity can be given

v(t) = 20-10t m/s

v(-1) = 30 m/s

v(5) = -30 m/s

**********************
I want to know what each result represents, and what is the meaning of the minus sign in v(5) = -30 m/s
The formula gives positive values for up and negative values for down; the -30 m/s just means it is traveling at 30 m/s in the down direction.
**********************
.............................

(d) Suppose that instead of being thrown from a stand the body was thrown vertically from the ground with the same acceleration and initial velocity . When will the body strike the ground and find its velocity then.

Here I changes the equation of height to be
h(t) =20t-5t^2

the equation of velocity (not changed)
v(t) = 20-10t m/s

the body will strike the ground when h(t) =20t-5t^2 = 0

t(20-5t) = 0

t=0 seconds and t = 4 seconds

**************************
I want to know what each result represents.
The formula only tells you when the object was at height zero, not when it lands. It started from ground level, and after going high in the air is finished at ground level. The start was time 0, the landing was 4 seconds later.

**************************

the velocity can be given v(t) = 20-10t m/s

v(0) = 20 m/s

v(4) = -20 m/s

**************************
I want to know what each result represents. and what is the meaning of the minus sign in v(4) = -20 m/s
As above -20 merely means 20 m/s in the down direction. v_o = 20 means it started at 20m/s but in the up direction.
**************************

All answers included above.

voltaire101 · Nov 29, 2011

Thank you very much : "Daniel" "PeterO"

It is clear now.

Log in or Sign up

A body thrown upwards from a stand of height

voltaire101

danielakkerma

PeterO

voltaire101

Thrown upwards from cliff (Replies: 5)

The max height the body reaches when this body is thrown verticaly upwards (Replies: 1)

What height was it thrown from. (Replies: 1)

Stones thrown from different heights (Replies: 1)

Trajectories of objects thrown from a height (Replies: 7)