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.

Question

perl · Answer

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

perl · Answer

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

anonymous · Answer

i dont know the last two vairables

anonymous · Answer

@perl

perl · Answer

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

perl · Answer

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

anonymous · Answer

y=-16t^2+72t

anonymous · Answer

@perl

anonymous · Answer

@rational

perl · Answer

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

anonymous · Answer

i dont know what those values are. @perl