If a ball is thrown vertically upward from the roof of a 64 foot tall building with a velocity of 16 ft/sec, its height in feet after t seconds is s(t)=64+16t-16t^2. What is the maximum height the ball reaches? What is the velocity of the ball when it hits the ground (height0)?
Please explain how you get the answers.
Please explain how you get the answers.
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v = ds/dt = 16 - 32t
Max height occurs when the velocity = 0 (the ball has stopped moving).
0 = 16 - 32t
t = 0.5 seconds
s(0.5) = 64 + 15(0.5) - 16(0.5)^2 = 67.5m < - - - MAX HEIGHT
Now, it will reach the ground when height = 0
0 = 64 + 16t - 16t^2
t = 2.562 seconds (ignore the negative solution of course)
So velocity when it reaches the ground, at 2.562 seconds, is
v(2.562) = 16 - 32(2.562) = -65.984m/s or 65.984m/s down
Max height occurs when the velocity = 0 (the ball has stopped moving).
0 = 16 - 32t
t = 0.5 seconds
s(0.5) = 64 + 15(0.5) - 16(0.5)^2 = 67.5m < - - - MAX HEIGHT
Now, it will reach the ground when height = 0
0 = 64 + 16t - 16t^2
t = 2.562 seconds (ignore the negative solution of course)
So velocity when it reaches the ground, at 2.562 seconds, is
v(2.562) = 16 - 32(2.562) = -65.984m/s or 65.984m/s down