Question

A 21.0-kg cannonball is fired from a cannon with muzzle speed of 900 m/s at an angle of 33.0° 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 hsecond ball (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- Esecond ball

Answer 1

(a) First we calculate h_first ball
Initial speed of first ball (V) = 900 m/s
launching angle ( $\theta$ ) = 33
Now we know that the maximum height is given by
H_max = (V²Sin² $\theta$ ) /(2g)
Therefore maximum height attained by the first ball
h_{first ball} = (900Sin33)²/(2*9.81) = 12246.262 m
Now for the second ball
h_{second ball} = (VSin90)²/2g = V²/2g = 900²/(2*9.81) = 41284.404 m
(b)
E_{first ball} = Potential energy + kinetic energy(due to horizontal component of velocity)
= mgh_{first ball} + (1/2)m(VCos33)² = 21*9.81*12246.262 + (1/2)*m*(900Cos33)²
= 2522852.435 + 5982147.575 = 8.505*10⁶ J
Now for the Second ball ,
Second ball willl do not have any velocity at top , therfore
E_{second ball} = mgh_{second ball} =21*9.81*(41284.404) = 8.505*10⁶ J