Question

A batter hits a softball straight up at 62 m/s how high up does it go?

Answer 1

physics problem: Initial velocity is 62 m/s and we want to find the max height. Final velocity at this height will be 0 m/s. The acceleration will be the negative force of gravity. Can you solve it now?

Answer 2

no...-9.8 is the force of accelleratiion due to gravity...not sure of the formula

Answer 3

where x(f)-x(0) = change is x or distance

Answer 4

v(f)^2=v(0)^2+2a(d)

Answer 5

uhhh..k...still not sure...anyone else?

Answer 6

Hopefully somebody can collaborate my formula given, but that is what I remember it to be. You know everything but d to solve the equation. Good luck

Answer 7

thanks

Answer 8

v² = u² + 2as

where v = final velocity
u = initial velocity
s = displacement
a = acceleration (in this case due to gravity)


Initially, velocity = 62m/s and
finally at maximum height, velocity = 0 m/s

v² = u² + 2as

where v = 0 m/s
u = 62 m/s
s = ?
a = (gravity) -9.81


Replacing values :

0 = 62² + (2 x s x -9.81)
62² = 2s x (9.81)
s = 62² / (2 x 9.81)
= 195.922528 m