Question

A 19.0-kg cannonball is fired from a cannon with muzzle speed of 1 150 m/s at an angle of 35.0° with the horizontal. A second ball is fired with the same initial speed at an angle of 90.0°. Let y = 0 at the cannon.

(a) Use the isolated system model to find the maximum height reached by each ball.

hfirst ball	=  m
hsecond ball	=  m

(b) Use the isolated system model to find the total mechanical energy of the ball–Earth system at the maximum height for each ball.

Efirst ball	=  J
Esecond ball	=  J

Answer 1

weight of canon ball = 19.0-kg

muzzle speed =1150 m/s

angle of 35.0° with the horizontal.

A second ball is fired with the same initial speed at an angle of 90.0°.

Let y = 0 at the cannon.

part A

1) Height of the first ball

where

is final velocity (at max height )

is muzzle speed  ; = 1150 m/s

is the angle

at max height velocity will be 658.95 m/s

we know that

where

is kinetic energy

is mass

is velocity

and

where

is potential energy

is mass , is acceleration due to gravity and h is height

at maximum height

PE=KE

so

here M=m

then   

put values we get

Max height reached by first ball is 22153.83 m.

1) Height of the second ball

here

sin 90⁰ = 1

at max height velocity will be 1150 m/s

here also

put values , we get

Max height of the second ball will be 67474.48 m.

part B

total mechanical energy is equal to the initial kinetic energy

which is

for first ball

J

for second ball

Total energy for both balls will be 12563750 J