A tennis player hits a ball at ground level, giving it an initial velocity of 24

Question

A tennis player hits a ball at ground level, giving it an initial velocity of 24.0 m/s at 57.0 degrees above the horizontal.

a.What are the horizontal and vertical components of the ball's initial velocity?
b. How high above the ground does the ball go?
c. How long does it take the ball to reach its maximum height?
d.What is the ball's velocity at its highest point?
e.What is the ball's acceleration at its highest point?
f. For how long a time is the ball in the air?
g.When this ball lands on the court, how far is it from the place where it was hit?

please show all steps...
thank you!

Explanation / Answer

a) v(x) = 24m/s cos57° = 13m/s v(y) = 24m/s sin57° = 20m/s b) v² = v0² + 2g?y Solved for ?: ?y = (v² - v0²) / 2g At the top of the flight path, the speed (v)of the ball is zero (vertically), so: ?y = [0 - (20m/s)²] / (2 x -9.8m/s²) = 20m c) ?y = (v0 + v)t / 2 Solved for t: t = 2?y / (v0 + v) = (2 x 20m) / (20m/s + 0) = 2.0s d) At the highest point, the balls's acceleration is 9.8m/s², downward. It's speed is 24m/s, horizontally, 0 vertically. e) Since it took 2.0s to reach its high point, it will take an additional 2.0s to come down, so 4.0s. f) ?x = (v0 + v)t / 2 = (13m/s + 13m/s)(4.0s) / 2 = 52m

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