PROJECTILE MOTION

Question

During Raptors vs Warriors game, with the latter in possession the score is deadlock at 70 – all. With 1.5 seconds remaining on the game clock, Stephen Curry (1.780 m high) decides to shoot. If he is 4.5 m away from the ring (3.048 m high)

a. At what angle must he release the ball in order to score if he shoots at 7.8 m/s?

b. If Kawhi Leonard with standing reach (2.91 m high) is standing 1 m away from the ring jumps in order to block the shot, how high must he jump to force an overtime?

Answer 1

This is me doing the problem. I don't know if they want you to assume the ball will reach the net in 1.5 seconds, but I think that would be an inaccurate assumption.

d_x = v_oxt = 7.8cos(0)t = 4.5 m (he is 4.5 m from hoop.

d_y = v_oyt + 1/2*a*t² = 7.8sin(0) +.5(-9.8)t² = 3.048 - 1.780 = 1.268 (difference in height between hoop and Steph Curry.

I now have two equations with two unknowns.

If I solve the first equation for 0, 0 = cos^-1(4.5/(7.8t))

solving second equation for 0, I get 0 = sin^-1((1.268+4.9t²)/(7.8t)

Now I can graph these two lines on a graphing calculator. Since t is my x, I will set x window from 0 to 5.

Since 0 is my y, I will set y window from 0 to 90 and use degree mode.

Solve the simultaneous equations, I get t₁ = 0.826s and 0₁ = 45.7^o

I also get a second solution t₂ = 1.15s and 0₂ = 60.0^o

Now to calculate height.

If Kawhi is 1m from hoop and Steph Curry is 4.5 m from hoop. That means we care about the height of the ball d_y when d_x = 4.5 -1 = 3.5m.

d_x = 7.8cos(0)t = 7.8*cos(45 or 60) t = 3.5

i decided just to graph in parametric mode now. using x = 7.8cos(45 or 60) t

y = 1.780 + 7.8sin(45 or 60)t + 0.5 (-9.8) t²

For either 0, I get goaltending. Ball is on its way down. Kawhi Leonard cannot block the shot and must hope the ball does not go in.

Answer 2

a) Use the kinematic equation Δx or Δy = v_o t + 1/2 a t². I'm assuming a = -g = -9.8 m/s². Δx is given as 4.5 m, while Δy is the difference 3.048 – 1.78 = 1.268 m. v_o is the appropriate component of initial velocity v = 7.8 m/s. The unknown is the launch angle @.

Δx = 4.5 = (7.8 cos @) t + 0 (no horizontal acceleration) so t can be written as 4.5/(7.8 cos @).

Δy = 1.268 = (7.8 sin @) t - 4.9 t² and substituting for t from above yields 1.268 = (4.5)(7.8 sin @)/(7.8 cos @) – (4.9)(4.5²)/(7.8² cos² @).

Multiply through by cos² @: 1.268 cos² @ =  4.5 sin@ cos@ - (4.9)(4.5²)/(7.8²). My algebra then failed me, but I quickly found that @ = 45.7 degrees is a solution.

b) Now Δx = 3.5 = (7.8 cos 45.7) t, and t = 0.6425 s.

Δy =  (7.8 sin 45.7) t - 4.9 t² = 3.587 – 2.023 = 1.564, so for total height y we add 1.78m to get 3.34 m when x is 3.5 m. Kawhi would have to jump at least 0.43 m to make up the difference and touch the ball.