A tennis player standing 12.6 m from the net hits the ball at 2.97° above the horizontal. To clear the net, t?
I do
vi=\sqrt(2gh/sin^2(theta)) and get 16.088
but for some reason it's wrong. Why?
Update:A tennis player standing 12.6 m from the net hits the ball at 2.97° above the horizontal. To clear the net, the ball must rise at least 0.385 m. If the ball just clears the net at the apex of its trajectory, how fast was the ball moving when it left the racket?
More Questions From This User See All
Answers & Comments
Verified answer
Equations are :
1)
Vy^2 = 2gh = 19,6*0.385 = 7.55 m2/sec^2
Vy = √7.55 = 2.75 m/sec
sen 2.97° = 0.0518
Vy = V1*sen 2.97° = 2.75 m/sec
V1 = Vy/sen 2.97° = 2.75/0.0518 = 53,0 m/sec approx.
2)
12.6 = V2^2/2g*sin 5.94° = V2^2/19.6*0,1035
V2^2 = 12.6*19.6/0.1035 = 2386 m^2/sec^2
V2 = √2386 = 49.0 m/sec approx.
as u can see the computed speed on both axes (V1 = 53 and V2 = 49) are not coincident ...this means that given data are not completely consistent ....the easier approach is that on x axis, based on the fact that apex of trajectory divides the full trajectory into two symmetrical parts ....so :
t = D/V2x = 12.6/(V2*cos2.97°) = 12.6/49 = 0.257 sec ( cos 2.97 is 1.00)
Vy = 49*sin 2,97 = 49*0.0518 = 2.54 m/se
h = Vy*t-4,9t^2 = 2.54*0.257 -4.9*0.257^2 = 0.330 m approx.
if we replace 0,385 with 0.330 in equation 1), then :
Vy^2 = 2gh = 19.6*0.330 = 6.468 m2/sec^2
Vy = √6.468 = 2.54 m/sec
V1 = 2.54/sin 2.97 = 2.54/0.0518 = 49.0 m/sec
so...given h is wrong !!!
Range=25.2 m Max H=0.330 m there fore u in y direction=6.468 and u in x direction=19.481 hence initial vel=20.527 m/s