Asked by Alex
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?
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?
Answers
Answered by
Henry
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.
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.
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