a. Vo = 12m/s[8o]
Xo = 12*cos8 = 11.88 m/s.
Yo = 12*sin8 = 1.67 m/s.
Tr = -Yo/g = -1.67/-9.8 = 0.170 s. = Rise time.
h = ho + (Y^2-Yo^2)/2g
h = 1.67 + (0-(1.67^2))/-19.6 = 1.74 m.
Above gnd.
d = 0.5g*t^2 = 1.74-1.0
4.9*t^2 = 0.74
t^2 = 0.151
Tf = 0.389 s. = Fall time.
Dx = Xo * (Tr+Tf)=11.88 * (0.170+0.389)=
6.64 m = Hor. distance. So it falls short of the net.
b. Missed by: 7-6.64 = 0.36 m.
A tennis player hits a ball 1.6 m above the ground. The ball leaves
the racket with a speed of 12 m/s at an angle 8.0 degrees above the horizontal. The net is a horizontal distance of 7.0 m away from the player and is 1.0 m high.
(a.) Does the ball clear the net? Explain how you determined this. You must do a
correct numerical calculation as part of your explanation.
(b.) If the ball clears the net, by what distance does it clear it? If not, by what distance
does it miss?
1 answer