Question

With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? note negligible resistance

Answer 1

Given

a = -9.8 m/s^2

vf = 0 m/s

d = 91.5 m

To Find

vi = ??

t = ??

Solution

First, find speed in units of m/s:

vf^2 = vi^2 + 2*a*d

(0 m/s)^2 = vi^2 + 2*(-9.8 m/s^2)*(91.5 m)

0 m^2/s^2 = vi^2 - 1793 m^2/s^2

1793 m^2/s^2 = vi^2

vi = 42.3 m/s

Now convert from m/s to mi/hr:

vi = 42.3 m/s * (2.23 mi/hr)/(1 m/s)

vi = 94.4 mi/hr

Answer 2

Answer:

= 0^2-u^2=2*(-9.8)*91.5

= -u^2=(-1793.4)

= u=under root 1793.4

= 42.35m/s

= 42.35*2.23mil/sec

= 94.44 mil/sec