A projectile of mass 5.0 kg is shot

Question

A projectile of mass 5.0 kg is shot horizontally with an initial speed of 17 m/s from a height of 25.0 m above a flat desert surface. For the instant before the projectile hits the surface, calculate each of the following: 

a) the work done on the projectile by gravity (1.23x103 J)

b) the change in kinetic energy since the projectile was fired. (1.23x103 J)

c) the final kinetic energy of the projectile. (1.9x103 J)

d) the final speed of the projectile (28 m/s)

Answer 1

v_ix = 17 but v_iy = 0 (initially, no movement in the y-direction, just in the x-direction). vfy is calculated just like if the object was dropped so, using one of the kinematic equations v_fy² = v_iy² + 2g(y_f - y_i) or v_fy = √(0 + 2(-9.81)(-25) = 22.147 m/s. But v_fx is still the same as v_ix because there are no forces in the x-direction so v_fx = 17 m/s. The speed is the vector addition of those two (calculated by using the Pythagorean Theorem) so v_f² = .v_fy² + v_fx² or v_f = .√ (v_fy² + v_fx²) = √ (22.147² + 17²) or vf = 27.92 m/s and using 2 sig figs, that rounds to 28 m/s

The work done by gravity is the force of gravity times the distance moved in the direction of that force (straight down) [ W = Fd]. The force of gravity is the weight or mg which is (5.0)(9.81) = 49.05 N and the distance in the y direction is 25 m so the work done is (49.05)(25) = 1226.25 J and using 2 sig figs, it is rounded to 1200 J.

The initial Kinetic Energy is 1/2mv_ix². The final KE is 1/2m_vf² so the change in KE is 1/2m(v_f² - v_ix²) = 1/2(5)(27.92² - 17²) = 1226.25 J or, rounding to two sig figs, 1200 J. The final KE of the projectile is again 1/2m_vf² = 1/2(5)(27.92²) = 1948.75 J or rounded to two sig figs, 1900 J