Projectile motion of a tennis ball

[Idealism_Theory]

Homework Statement

A tennis player hits a ball at ground level, giving it an initial velocity of 21.0 m/s at 58.0 deg above the horizontal. What is the ball's velocity at the highest point?

Homework Equations

Vx=Vocos(theta)
Vy=Vosin(theta)
y=1/2-g(t^2)+Vy(t)
V-Vo/a = t

The Attempt at a Solution

I've completed all of the steps up to this point and arrived at the conclusion that the velocity of the ball at the highest point must be zero, theoretically speaking, of course. When I input 0 m/s, the answer was marked as incorrect. What am I missing. Final velocity must be zero since the balls velocity slows down to zero when it's at the maximum distance. It also asked for the acceleration at the highest point and theoretically speaking, the answer was 9.8 m/s^2, which is consistent with theory. What gives?!

 

[atavistic]

Hehe stumped arent you?
But think.There are 2 velocities.Vx and Vy.Which one is zero?Both or one of those?

 

[Idealism_Theory]

Hehe stumped arent you?
But think.There are 2 velocities.Vx and Vy.Which one is zero?Both or one of those?

Yes, I am completely stumped. If they're asking for velocity at highest point, naturally I'd think Vy. I didn't think Vx could even have a "highest" point?! Only Vy has a velocity of zero. As for Vx, what could what could that possibly be?

Here's what I'm thinking now...

Using equation v = Vo + at

Substitute initial velocity of x (Vx) which is 11.1m/s, the acceleration of 9.8 m/s^2 and the time it takes to reach the balls max heigh which is 1.82s ...

so, v = 11.1m/s + 9.8m/s^2(1.82s)
v = 28.9 m/s

Is that correct, or am I fooling myself? Doesn't make sense to me!

 

[atavistic]

NO.You got the question wrong.They asked the V and not the Vy.At any point V is given as under root Vy^2 + Vx^2.

And what is the equation u put above.Makes no sense.you can't combine Vx with Ay.Vx is always constant.Only Vy changes with time.

 

[Feldoh]

Yes, I am completely stumped. If they're asking for velocity at highest point, naturally I'd think Vy. I didn't think Vx could even have a "highest" point?! Only Vy has a velocity of zero. As for Vx, what could what could that possibly be?

Here's what I'm thinking now...

Using equation v = Vo + at

Substitute initial velocity of x (Vx) which is 11.1m/s, the acceleration of 9.8 m/s^2 and the time it takes to reach the balls max heigh which is 1.82s ...

so, v = 11.1m/s + 9.8m/s^2(1.82s)
v = 28.9 m/s

Is that correct, or am I fooling myself? Doesn't make sense to me!

Vo doesn't equal Vx

 

[Idealism_Theory]

Vo doesn't equal Vx

Well, does Vo the originally stated 21.0m/s? That would mean the final velocity is 38.8 m/s... still not making sense. Please clarify, someone!

Originally Posted by Idealism_Theory
Yes, I am completely stumped. If they're asking for velocity at highest point, naturally I'd think Vy. I didn't think Vx could even have a "highest" point?! Only Vy has a velocity of zero. As for Vx, what could what could that possibly be?

Here's what I'm thinking now...

Using equation v = Vo + at

Substitute initial velocity of x (Vx) which is 11.1m/s, the acceleration of 9.8 m/s^2 and the time it takes to reach the balls max heigh which is 1.82s ...

so, v = 11.1m/s + 9.8m/s^2(1.82s)
v = 28.9 m/s

Is that correct, or am I fooling myself? Doesn't make sense to me

 

[Idealism_Theory]

Vo doesn't equal Vx

NO.You got the question wrong.They asked the V and not the Vy.At any point V is given as under root Vy^2 + Vx^2.

And what is the equation u put above.Makes no sense.you can't combine Vx with Ay.Vx is always constant.Only Vy changes with time.

Well, I tried that. It was incorrect! Now what do you suggest. I definitely didn't see that as the answer either.

v = sqrt vx^2 + vy^2

v = sqrt 11.1^2 + 17.8^2

v = 30 m/s (which is marked as incorrect)

 

[Idealism_Theory]

Bump.

 

[atavistic]

Put Vy = 0 man. so that gives u 11.1 m/s as the answer.