Question
A tennis ball is released at 30 degrees. It reaches max height of 20 ft find initial speed
Answers
bobpursley
So we are to assume its initial velocity was 30 deg above the horizon.
verticalspeed=vsin30
The KE in the horizontal direction remains constant, but in the vertical
1/2 m v'^2+mgh=1/2 m (V sin30)^2
at the top, v' is zero, so
g*20=1/2 V^2 *1/4)
v^2=8g*20ft=160ft*32ft/s^2
v=sqrt (160*32) ft/sec=21.6ft/s
check that.
verticalspeed=vsin30
The KE in the horizontal direction remains constant, but in the vertical
1/2 m v'^2+mgh=1/2 m (V sin30)^2
at the top, v' is zero, so
g*20=1/2 V^2 *1/4)
v^2=8g*20ft=160ft*32ft/s^2
v=sqrt (160*32) ft/sec=21.6ft/s
check that.
Henry
Y^2 = Yo^2 + 2g*h = 0.
Yo^2 - 64*20 = 0, Yo = 35.8 Ft/s.
Vo*sin30 = 35.8, Vo = 71.6 Ft/s.
Yo^2 - 64*20 = 0, Yo = 35.8 Ft/s.
Vo*sin30 = 35.8, Vo = 71.6 Ft/s.