Question 1178129: a person throws a ball upward into the air with an initial velocity of 15.0 m/s.
a) calculate how high it goes
b)calculate how long the ball is in the air before it comes back to his hand.
c) how much time it takes for the ball to reach the maximum height?
d) find the velocity of the ball when it returns to the thrower's hand
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website!
Here acceleration (a) is g = 9.8m/s^2 (this will be negative since it is going in the opposite direction of the initial velocity),
initial velocity(u) = 15m/s
final velocity (v) = 0m/s if the ball stops
v^2 = u^2 + 2a*h, where h is the max height of the ball,
we can find h by solving: h = (v^2 - u^2)/2a,
we plug in h = (0^2 - 15^2)/2(-9.8) = 11.48 m.
b)calculate how long the ball is in the air before it comes back to his hand.
time to reach its peak and back down to our hand.
Using this, the time it takes to get to the top * 2 = total time of flight . Using d = (v + u)*t / 2, ) = 2(11.48)/(15) = 1.53 s to reach top,
we get a total of2*1.53 = 3.06 s to go up and back down to our hand.
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