How Do You Calculate the Initial Velocity Components of a Projectile?

[kara]

2 seconds after being projected from ground level, a projectile is displaced
40 m horizontally and 53 m vertically above its launch point. What are the horizontal and vertical components of the initial velocity of the projectile? At the instant the projective ahieves maximum height above ground level, how far is it displaced horizontally from the launch point?

Any suggestions for that questions, I've already drawn a diagram.

 

[Doc Al]

Start by writing equations that describe the horizontal and vertical displacement as a function of time.

 

[kara]

so would saying the following make any sense:

the horizontal velocity is constant so the projectile is moving 40m/s right
the vertical velocity changes 9.80 m/s downward because of gravity

 

[Doc Al]

the horizontal velocity is constant so the projectile is moving 40m/s right

The horizontal velocity component is constant, but it's not 40 m/s.

the vertical velocity changes 9.80 m/s downward because of gravity

The vertical velocity component accelerates (due to gravity) at 9.8 m/s^2 downward; acceleration is the rate of change of velocity (m/s per s).

 

[kara]

how do i find the horizontal velocity component?

 

[Doc Al]

Use the basic definition of velocity.

 

[kara]

can i do this:

y = 1/2 * g* t^2
y = 1/2 (9.8) (2)^2
y = 19.6

 

[kara]

or is that only when an object is thrown directly upwards and falls directly downwards.

 

[kara]

alright, i know that v = d/t so i'd get v = 40m/2sec = 20 m/sec but do i use the horizontal distance or vertical distance ? and what does velocity give me? is that the horizontal component?

 

[Doc Al]

Well... 40m is the horizontal distance... so that must be the horizontal component of velocity.

 

[kara]

now how do i find vertical component?

 

[Doc Al]

Do what I suggested in post #2. What equation describes the vertical motion?

 

[kara]

y = y0 + v0y*t - 1/2*g*t^2 ?

 

[kara]

so:

- initial position (y0) would be 0 m
- initial velocity (v0) would be 0 m/s
- time would be 2.0 sec
- force of gravity 9.8 m/s^2

so my answer would be -19.6 m/s??

 

[kara]

wait.. wouldn't gravity be -9.8 m/s^2 so that the answer would be 19.6 ?

and i don't know what units it'd be

 

[BishopUser]

what makes you think its initial velocity is 0

 

[kara]

its on the ground?

 

[BishopUser]

The velocity you calculated is if you dropped it off a cliff (in which case its initial velocity would be zero). However, in this case it is projected upwards meaning it was given some initial velocity

 

[Doc Al]

so:

- initial position (y0) would be 0 m

OK.

- initial velocity (v0) would be 0 m/s

v0 is what you are trying to find! Don't assume it's anything--certainly not zero. You know it moved up so v0 can't be zero!

- time would be 2.0 sec

OK

- force of gravity 9.8 m/s^2

The acceleration due to gravity is -9.8 m/s^2.

You left out one piece of data: the vertical displacement at time t.

Solve for v0.

 

[BishopUser]

OK.

v0 is what you are trying to find! Don't assume it's anything--certainly not zero. You know it moved up so v0 can't be zero!


OK

The acceleration due to gravity is -9.8 m/s^2.

You left out one piece of data: the vertical displacement at time t.

Solve for v0.

take into consideration his equation has a built in negative so if he put g in as negative it would make the object fly upwards

 

[Doc Al]

take into consideration his equation has a built in negative so if he put g in as negative it would make the object fly upwards

Right... don't put g as negative! g is always a positive number; the acceleration due to gravity is -g. (Which is why your equation has the minus sign already.)

 

[kara]

so all the - demonstrates is the negative direction?

 

[kara]

you said i left out the vertical displacement at time, so that would be y in the case of this equation y = y0 + v0y*t - 1/2*g*t^2 , and y is 53 m ?

 

[Doc Al]

so all the - demonstrates is the negative direction?

Sure, but that makes a lot of difference. The generic kinematic formula (for constant acceleration) is:
y = y0 + v0t + 1/2at^2

When the acceleration is due to gravity (like this problem) we put a = -g:
y = y0 + v0t -1/2gt^2

 

[Doc Al]

you said i left out the vertical displacement at time, so that would be y in the case of this equation y = y0 + v0y*t - 1/2*g*t^2 , and y is 53 m ?

Exactly...

 

[kara]

you said i left out the vertical displacement at time, so that would be y in the case of this equation y = y0 + v0y*t - 1/2*g*t^2 , and y is 53 m ?

 

[kara]

alright so solving that equation would give me v0. Now what do i do with that? would i also need to solve for v0 in equation x=x0 + v0t?

 

[kara]

:O I just figured something out... Don't know if I am on the right track but let me know:

If i solve for v0 in equation x=x0 + v0t, then i can sub that into the other equation?

 

[Doc Al]

Solving that equation gives you the vertical component of the initial velocity; you already solved for the horizontal component.

 

[kara]

Alright, so i got 36.3 m/s for v0

 

[kara]

So the horizontal component of my initial velocity is 20 m/s, and the vertical component of initial velocity is 36.6 m/s

 

[kara]

Now i have to calculate how far is its displacement horizontally from launch pt, at the instant it achieves max height.

So i know that its v will be 0 m/s at max heigh.

 

[Doc Al]

So i know that its v will be 0 m/s at max heigh.

Right... the vertical component of the velocity will be zero.

 

[kara]

and b/c I am looking for how far its been displaced horizontally from the launch pt. i am looking for x?

 

[kara]

so i can solve for t in the y=y0 + v0-1/2gt^2 equation and sub t into x=x0 +v0t equation and solve for x

 

[Doc Al]

and b/c I am looking for how far its been displaced horizontally from the launch pt. i am looking for x?

That's right. You are looking for the value of x when y is maximum. Hint: When does it reach the maximum height?

 

[kara]

well the max height is 53 m, and it reaches that height when v = 0.0 m/s

 

[kara]

i plugged in all my values but get stuck at one point with a negative square root:

53 = -1/2(9.8)t^2

 

[kara]

if i multiply both sides by 2 to get rid of the 1/2 i get

106 = -(9.8) t^2

 

[Doc Al]

i plugged in all my values but get stuck at one point with a negative square root:

53 = -1/2(9.8)t^2

You left out part of that equation; it should be:
53 = v0t -1/2(9.8)t^2

Where v0 is the vertical component of initial velocity that you found earlier.