Projectile motion Q (1 Viewer)

[Adrian.]

Hi. I have a projectile motion question I am having trouble with.

1. A tennis player tosses the ball into the air to serve. It takes 2.2s between tossing and serving. Assume that the racquet strikes the ball at the same height from which it was tossed.

a) calculate the speed with which the ball was tossed.

u = 5.39 (I got this right)

b) Calculate the height achieved, above the player's hand, by the ball.

This one I had trouble with.

y=ut + 1/2at²
y = 5.39 X 1.1 + 1/2 X -9.8 X 1.1²
y = 5.929 - 5.929
y = 0

In the answers they have done this.

y = 5.39 X 1.1 + 1/2 (-9.8)(0.55)²

= 1.48m

Why have they halved the time again?

Thank you.

-Edit- HTML doesn't work

 

[PookieMonster]

its all aboot ze vomit.

 

[Adrian.]

Thank you for that!

 

[dangerousdave]

You have half time because thats when its at maximum height. The answer should've read:
y = 5.39 X 0.55 + 1/2 (-9.8)(0.55)2
y = 1.48m

edit: ok i read the question again, i have no idea what's going on.

 

[KFunk]

It's 2.2 seconds up and back, so that's 1.1 seconds to reach the peak. I think there's something funky going on with your question.

Actually, i'm unsure about your initial velocity since:

v = u +at
0 (velocity at max. height = 0) = u + (-9.8)x(1.1)

i.e u = 9.8x1.1 = 10.78 = double your velocity??

 

[Adrian.]

OK, I'll show my working to a.

y = ut + 1/2at²

Bingo, we don't know y. I put it as zero. Funny though the answers have made the same mistake.

So if we use
v = u + at
0 = u - 9.8 X 1.1
u = 10.78

Going onto the next part

b)
y = ut + 1/2at²
y = 10.78 X 1.1 + 1/2 X -9.8 X 1.1²
y = 5.929m

This seems like a damn high 'toss' to me. Damn this question is driving me nuts.

 

[AntiHyper]

The question should also said that the tennis player tossed the ball straight up.

 

[Adrian.]

The question should also said that the tennis player tossed the ball straight up.

But that doesn't change anything does it?

 

[rama_v]

OK, I'll show my working to a.

y = ut + 1/2at²

Bingo, we don't know y. I put it as zero. Funny though the answers have made the same mistake.

So if we use
v = u + at
0 = u - 9.8 X 1.1
u = 10.78

Going onto the next part

b)
y = ut + 1/2at²
y = 10.78 X 1.1 + 1/2 X -9.8 X 1.1²
y = 5.929m

This seems like a damn high 'toss' to me. Damn this question is driving me nuts.

I think u got the question right, I tried it and got the same. Obviously they made some sort of an error in the book...