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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).

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.

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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.

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(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

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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

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

## Homework Statement

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.

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

## Homework Equations

## 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

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

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