Question

A pitcher throws a baseball straight into the air with a velocity of 72 feet/sec. If acceleration due to gravity is −32 ft/sec2, how many seconds after it leaves the pitcher's hand will it take the ball to reach its highest point? Assume the position at time t = 0 is 0 feet.

Answer 1

You can use the equation:
$$ \Large {
y = -\frac g 2 t^2 + v_ot + y_o

}$$

Answer 2

v_0 = velocity at time zero
y_o = height at time zero
g = acceleration due to gravity

Answer 3

i dont know the last two vairables

Answer 4

@perl

Answer 5

we are given that v_o = 72 ft/s, and y_0 = 0

Answer 6

You can use the equation:
$$ \Large {
y = -\frac g 2 t^2 + v_ot + y_o
\\ \therefore \\
y = -\frac{32}{2}\cdot t^2 + 72t + 0

}$$

Answer 7

y=-16t^2+72t

Answer 8

@perl

Answer 9

@rational

Answer 10

yes thats correct. now find the maximum of that .
that occurs when t = -b / (2a)

Answer 11

i dont know what those values are. @perl