Question

A shell is launched with an initial velocity at an angle of 40.0° above horizontal

from ground level. The shell needs to hit 1.5 km away. There is no appreciable air resistance,
and g = 9.80 m/s2
.
a. What should the initial velocity be?
b. What are the components of the shell’s velocity when it hits the ground? (horizontal
and vertical components)

Answer 1

122.17 m/s

Step-by-step explanation:

x cos 40 = horizontal velocity

1500 m / x cos 40 = time in the air = 1958.11 / x

x sin 40 = vertical velocity

find when shell vertical velocity = 0 (this is max height....1/2 way through its flight) , the time when it hits the ground will be twice this...

0 = x sin 40 - 9.8 t

t = x sin40 / 9.8 time in the air is twice this = .13118 x

Equate the two times from above to solve for x

1958.11/ x = .13118 x

x = 122.17 m/s