karen hits a tennis ball with an initial velocity of 42 feet per second an at angle of 16 degree with the horizontal from a height of 2 feet. she is 20 feet from the net and the net is 3 feet. will the ball go over the net?

Answer 1

feet, yuuk :(
oh well, g = 32 ft/s^2 in that case

Vi = 42 sin 16 = 11.6 ft/s
u = 42 cos 16 = 40.4 ft/s

Horizontal problem:
goes 20 ft at 40.4 ft/s
t = 20/40.4 = .495 s

Vertical problem with t = .495 s
-m g = m d^2h/dt^2
d^2 h/dt^2 = -g = -32
dh/dt = Vi - 32.2 t
h = Hi + Vi t - (1/2)(32)t^2
h = Hi + Vi t - 16 t^2
h = 2 + 11.6(.495) - 16(.495)^2
= 3.82 feet, clears easily