Asked by Lynn
In a volleyball a player on one team spikes the ball over the net when the ball is 10 ft above court. The spike drives the ball downward with initial vertical velocity of -55 feet per second. Players on the opposing team must hit the ball back over the net before it touches the court. How much time do the opposing players have to get ready to hit the spiked ball?
I know that the formula is h=-16t^2+v0t+h0
The only thing I know what to do is...
10=-16t^2-55 ? ?
I know that the formula is h=-16t^2+v0t+h0
The only thing I know what to do is...
10=-16t^2-55 ? ?
Answers
Answered by
Reiny
Your formula would be
h = -16t^2 - 55t + 10
when it hits the ground h = 0
16t^2 + 55t - 10 = 0
t = (-55 ± √3665)/32 = .173 seconds or a negataive
h = -16t^2 - 55t + 10
when it hits the ground h = 0
16t^2 + 55t - 10 = 0
t = (-55 ± √3665)/32 = .173 seconds or a negataive
Answered by
David
10ft
Answered by
David
set up as a quadratic then solve for t
0=(8.98)t^2+55t-10
t=0.1767 seconds
0=(8.98)t^2+55t-10
t=0.1767 seconds
Answered by
Alex
Where did you get 16
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