Question

A ball is shot from the ground straight up into the air with initial velocity of 42 ft/sec. Assuming that the air resistance can be ignored, how high does it go? The acceleration due to gravity is 32 ft per second squared.

Would you please solve this question using CALCULUS, NOT physics equations?

The correct answer is 26.5625 ft.

Answer 1

dy/dt = v
dv/dt= -a
integrate
v = -at + A A= constant
42 = -a (0) + A
so A = 42

v=0 when
a t = 42 so
t = 42/32

integrate v
y = integral v dt = -(1/2) a t^2 + 42 t + B B another constant
t=0 , s = 0 grounded
0 = 0 + 0 + B
B=0

height at t =42/32 is

y = -(1/2)(32)(42/32)^2 + 42(42/32) + 0
y = 27.5 ft at t =42/32 and v = 0



s

Answer 2

Not sure why my number is a bit different from the answer. I keep getting 27.56.