What are the Kinematics of a Ball in Projectile Motion?

[timnswede]

Homework Statement

In a friendly game of handball, you hit the ball essentially at ground level and send it toward a wall 4.6 m away. The ball is hit with a speed of 20 m/s at 40 degrees above the horizontal.

a.) At what height does the ball strike the wall?

b.) Calculate the speed of the ball at impact.

c.) Is the ball moving upward or downward at impact?

2. The attempt at a solution

a.) I found the time it takes for the ball to hit the wall first: 4.6=20cos40t t=.3 s
Then I used that to find the height of the ball at that time. y=20sin40*(3)-4.9(.3^2) y=3.42m

b. I am not entirely sure how to do this one, but here is my attempt.
Horizontal velocity stays constant so that equals 20cos40 m/s.
Vertical velocity I calculated as Vy=20sin40-9.8(.3) Vy=9.9 m/s.
Then V=sqrt[((20cos40)^2)+(9.9^2)]= 18.2 m/s

c. For this I calculated what the maximum height of the ball would be.
0=(20sin40)^2 -9.8x
x=16.9 m which is higher than the 3.42 m I calculated earlier, so the ball is moving upwards.

 

[haruspex]

[
Horizontal velocity stays constant so that equals 20cos40 m/s.
Vertical velocity I calculated as Vy=20sin40-9.8(.3) Vy=9.9 m/s.
Then V=sqrt[((20cos40)^2)+(9.9^2)]= 18.2 m/s

That's certainly the right method. Haven't checked the num bers in detail but they look reasonable.
One trap to avoid is using a computed answer at one step, to only 1 or 2 decimal places, as input to a calculation in a later step. You get accumulation of errors. Better to keep everything in symbolic algebra as long as possible. Sometimes you get cancellations in later parts of the question, helping to reduce numerical error.

c. For this I calculated what the maximum height of the ball would be.
0=(20sin40)^2 -9.8x
x=16.9 m which is higher than the 3.42 m I calculated earlier, so the ball is moving upwards.

A bit more obvious is just to look at the sign of Vy in your working above.