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# Projectile Motion Numerical

## Question

Q:A tennis ball is served horizontally from 2.4m above the ground at 30m/s

(a)The net is 12m away and 0.9m high, will the ball clear the net?

(b)Where will the ball land?

:sorrow:

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Data:

V0x= 30m/s

H1= 2.4 m

V0y= 0

X= 12m

H2= 0.9m

To find:

(a) Will the ball clear the net?

( X’= ?

Solution:

(a) X= V0x t

12=30 t

12/30 = t

t= 0.4 s

As, H=V0y t + ½ g t2

H= (0)(0.4)+1/2(-9.8)(0.4)2

H= -0.78m

-ve sign shows that the height of the ball at the net from the top.

Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m

As 1.62m>0.9m so the ball will clear the net.

( H1= V0y t’ + ½ g t’2

-2.4= (0)t’ + ½ (-9.8) t’2

.

.

.

t’= 0.69s

X’=V0x t’

X’=(30)(0.96)

X’= 20.7m

This is the solution my teacher came up with and honestly I have no idea whats going on? :/

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How can the height be -ve!? :/

It doesn't make sense...

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The negative sign on the height just indicates that the ball is lower than when it started. Your teacher is calling the launch height 0, and determining when the ball crosses the net it has fallen that amount vertically. Check out our projectile motion tutorials/videos as a starting point and see if that helps. If not, if you can give us an idea of what you're thinking, what you've figured out so far, and where you're stuck, it will make providing assistance much more straightforward. Good luck!

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Q:A tennis ball is served horizontally from 2.4m above the ground at 30m/s

(a)The net is 12m away and 0.9m high, will the ball clear the net?

(b)Where will the ball land?

:sorrow:

This is a very good question and I got the same answer as your teacher :eagerness:

The first thing I should do is drawing the diagram to see the actual picture of the problem

Next, since this is Projectile Motion problem, this problem require 2 components, x and y

Now, no the given, so you can choose which equation to choose. For this, we choose the following equation [TABLE="class: maths"]

[TR]

[TD="align: right"]x = [/TD]

[TD="align: left"]x0 + v0Δt + ½aΔt2

[/TD]

[/TR]

[/TABLE]

Now, set the two components for x and y

y-component:

Yf - Yi = V(iy) t + 12 ayt2

You know the origin is where the ball is hit which makes yf=0 and yi which stand for y final equal to -2.4m. v initial is zero because the ball is throwing side away so imagine like dropping the ball. The acceleration is 9.8. Now sub it in, you get the amount of time that it takes to hit the ground.

x-component:

v initial is 30m/s

acceleration is 0 since the direction is going side way

just plug it in to the x component. After you have the equation set up, plug in the t which give you the answer for B. The distance that it hits the ground

For part A

just use 12m away, so you want to know when the ball flies pass at 12m.

Simple, use the x component equation. make x-xinitial=12m. Now you have the time which is 0.4second

The last step is plug into y component to find how high was the ball almost reach to the ground which is 0.784m

Now you can get the answer by simply take 2.4m-0.784m=20.7m

I hope this help, if you have trouble doing this, should check out more videos tutorials :star:

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Thanks for helping out our forum members!

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