Question

A 0.2 kg cannon ball is fired at an upward angle of 45° from the top of a 165 m cliff with a speed of 175 m/s. (A) Using conservation of energy, find the speed it has when it strikes the ground below. Suppose the cannon ball lands in a soft, muddy field. The force of the field is 75 N on the projectile. (B) How deep does it penetrate into the ground?

Answer 1

To solve this problem, it is necessary to apply the concepts related to the work done by a body when a certain distance is displaced and the conservation of energy when it is consumed in kinetic and potential energy mode in the final and initial state. The energy conservation equation is given by:

$\Delta KE_i + \Delta PE_i = \Delta KE_f + \Delta PE_f$

Where,

KE = Kinetic Energy (Initial and Final)

PE = Potential Energy (Initial and Final)

And the other hand we have the Work energy theorem given by

$\Delta KE = W = F*d$

Where

W= Work

F = Force

D = displacement,

PART A) Using the conservation of momentum we  can find the speed, so

$\Delta KE_i + \Delta PE_i = \Delta KE_f + \Delta PE_f$

$\frac{1}{2}mv_1^2 + mgh_i = \frac{1}{2}mv_f^2+mg_h2$

The height at the end is 0m. Then replacing our values

$\frac{1}{2}mv_1^2 + mgh_i = \frac{1}{2}mv_f^2+0$

Deleting the mass in both sides,

$\frac{1}{2}v_1^2 + gh_i = \frac{1}{2}v_f^2$

Re-arrange for find $v_f^2,$

$v_f^2= 2(gh_i)+v_1^2$

$v_f^2 = 2(9.8*165)+(175)^2$

$v_f=\sqrt{33859}$

$v_f = 184.008m/s$

PART B) Applying the previous  Energy Theorem,

$\Delta KE = W = F*d$

$\frac{1}{2}mv^2 = F*d$

$\frac{1}{2}(0.2)(184.008)^2 = (75)*d$

Solving for d

$d = 45.15 m$