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AS Physics Problem

Spinning a tennis ball allows it to be hit harder and still hit the table on the other side of the net.
A tennis ball is hit, without any spin, from one end of the table so that ti leaves the bat horizontally with a speed of 31 ms^-1. The length of the table is 2.7m.
Show that the ball falls a vertical distance of about 4cm as it travels the length of the table.
Find the time it takes to travel that displacement horizontally.
Then use suvats to find the vertical displacement.
Posted from TSR Mobile

I used s=ut+0.5t^2 and I got t= 0.74.But I did it for vertical and got s= 0.0268 cm for vertical displacement. Where's my mistake?
Reply 3
Avatar for kateyl
kateyl
3
Don't forget to think of projectile motion questions as two components: horizontal (s=d/t) and vertical (suvats).

1. using speed = distance/time, find the time taken for the ball to travel the length of the table. This time is also the same time taken for the ball to fall from its initial position vertically to the table's surface.

2. Now use s=ut+(1/2)at^2 to find the vertical displacement. You should get an answer that rounds up to 4cm.
Original post by kateyl
Don't forget to think of projectile motion questions as two components: horizontal (s=d/t) and vertical (suvats).

1. using speed = distance/time, find the time taken for the ball to travel the length of the table. This time is also the same time taken for the ball to fall from its initial position vertically to the table's surface.

2. Now use s=ut+(1/2)at^2 to find the vertical displacement. You should get an answer that rounds up to 4cm.

Thank you, followed your method and I got 0.039 which rounds to 0.04m or 4cm .
Reply 5
Avatar for kateyl
kateyl
3
Original post by Kadak
Thank you, followed your method and I got 0.039 which rounds to 0.04m or 4cm .


You're very welcome! :biggrin:
Original post by Kadak
Posted from TSR Mobile

I used s=ut+0.5t^2 and I got t= 0.74.But I did it for vertical and got s= 0.0268 cm for vertical displacement. Where's my mistake?

In case you didn't notice when you used s = vt, you missed the a term in your equation. It should be s=ut+12at2s = ut + \frac{1}{2}at^2
Original post by morgan8002
In case you didn't notice when you used s = vt, you missed the a term in your equation. It should be s=ut+12at2s = ut + \frac{1}{2}at^2

ooh, that might be why I didn't get it first time around . Ty

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