Question

1. A 26 kg cannon ball is fired from a cannon with muzzle speed of 962 m/s at an angle of 19.2 ◦ with the horizontal. The acceleration of gravity is 9.8 m/s 2 . Use conservation of mechanical energy to find the maximum height reached by first ball. Answer in units of m.2. A second identical ball is fired at an angle of 144◦ . What is the total mechanical energy at the maximum height of the second ball. Answer in units of J.

Answer 1

Answer

given,

Mass of the cannon ball = 26 Kg

speed of cannon with muzzle = 962 m/s

angle with horizontal = 19.2°

angle of second ball = 144°

for first ball

$v_y = v sin \theta_h$................(1)

$v_y = (962) sin 19.2^0$

$v_y = 316.37\ m/s$

now equating kinetic energy and potential energy

$\dfrac{1}{2}mv_y^2 = mgh$

$\dfrac{1}{2}v_y^2 = gh$

$h = \dfrac{v_y^2}{2g}$.................(2)

$h = \dfrac{316.37^2}{2\times 9.8}$

h = 5106.63 m

total mechanical energy of ball at maximum height

angle = 144° = (90° + 54°)

so θ = 54°

$E = \dfrac{1}{2}m(vsin\theta)^2$

$E =\dfrac{1}{2}\times 26 \times (962\times sin 54^0)^2$

E = 7.75 x 10⁶ J