# Help, Projectile Motion!

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#1
I'm slightly uselss at projectile motion questions! I can never apply the correct formulae to the given question! My exam is on thursday and i'm freaking out! For example which equation do i use to solve the height of the cliff if the horizontal speed is 22ms-1 and the time taken is 1.5s? i tried d=vt and s=ut+1/2at(t) but keep getting the wrong answer! any hints?
0
7 years ago
#2
(Original post by mirrk97)
I'm slightly uselss at projectile motion questions! I can never apply the correct formulae to the given question! My exam is on thursday and i'm freaking out! For example which equation do i use to solve the height of the cliff if the horizontal speed is 22ms-1 and the time taken is 1.5s? i tried d=vt and s=ut+1/2at(t) but keep getting the wrong answer! any hints?
Welcome to TSR.

We will be able to help you better if you post the whole of this question.
At the moment it's a bit too vague to give a definite answer.
Is the object actually projected horizontally from the top of the cliff or is that value the horizontal component?
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#3
(Original post by Stonebridge)
Welcome to TSR.

We will be able to help you better if you post the whole of this question.
At the moment it's a bit too vague to give a definite answer.

The Question is: A ball is kicked off a cliff with a horizontal speed of 22ms-1 the ball hits the ground 1.5s later. Calculate the height of the cliff.
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7 years ago
#4
(Original post by mirrk97)
The Question is: A ball is kicked off a cliff with a horizontal speed of 22ms-1 the ball hits the ground 1.5s later. Calculate the height of the cliff.
With all these questions you can consider the vertical motion separately from the horizontal.
If it's projected horizontally, it's initial vertical component of motion is zero.

So from the vertical motion point of view, it has zero initial velocity and it falls for 1.5s. It's acceleration is g.

So you have
s=?
u=0 m/s
a = 9.81m/s/s
t = 1.5s

So you are using the correct equation s = ut + 1/2 at2

If you are getting the wrong answer then
- are you subbing the correct values
- are you mis keying the calculation
0
#5
(Original post by Stonebridge)
With all these questions you can consider the vertical motion separately from the horizontal.
If it's projected horizontally, it's initial vertical component of motion is zero.

So from the vertical motion point of view, it has zero initial velocity and it falls for 1.5s. It's acceleration is g.

So you have
s=?
u=0 m/s
a = 9.81m/s/s
t = 1.5s

So you are using the correct equation s = ut + 1/2 at2

If you are getting the wrong answer then
- are you subbing the correct values
- are you mis keying the calculation
Thank You! I was using the wrong value of a this helped me especially knowing i was starting on the right foot! thank you!
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