A girls throws a tennis ball straight into the air with a velocity of 64 feet/sec. If acceleration due to gravity is −32 ft/sec2, how many seconds after it leaves the girl's hand will it take the ball to reach its highest point? Assume the position at time t = 0 is 0 feet.
2 seconds :)
2 seconds
To find the time it takes for the ball to reach its highest point, we can use the kinematic equation:
v = u + at,
where:
v = final velocity (the velocity at the highest point)
u = initial velocity (the velocity at t = 0)
a = acceleration
t = time
In this case, we want to find the time it takes for the ball to reach its highest point, so v = 0 ft/sec (since at the highest point, the ball momentarily stops moving before it starts to fall back down).
We are given:
u = 64 ft/sec
a = -32 ft/sec^2 (negative because acceleration due to gravity acts downward)
Using the kinematic equation, we have:
0 ft/sec = 64 ft/sec + (-32 ft/sec^2) * t.
Simplifying the equation:
0 ft/sec = 64 ft/sec - 32 ft/sec^2 * t.
Rearranging the equation to solve for t:
32 ft/sec^2 * t = 64 ft/sec.
Dividing both sides of the equation by 32 ft/sec^2:
t = 64 ft/sec / 32 ft/sec^2.
The units ft/sec cancel out, leaving:
t = 2 seconds.
Therefore, it will take the ball 2 seconds to reach its highest point after it leaves the girl's hand.
h(t) = 64t-16t^2
so, at max height, dh/dt=0.
So, find t for that.