William W. answered • 10/26/19

Experienced Tutor and Retired Engineer

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_{fy}^{2} = v_{iy}^{2} + 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_{f}^{2} = .v_{fy}^{2} + v_{fx}^{2} or v_{f} = .√ (v_{fy}^{2} + v_{fx}^{2}) = √ (22.147^{2} + 17^{2}) 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_{ix}^{2}. The final KE is 1/2m_{vf}^{2} so the change in KE is 1/2m(v_{f}^{2} - v_{ix}^{2}) = 1/2(5)(27.92^{2} - 17^{2}) = 1226.25 J or, rounding to two sig figs, 1200 J. The final KE of the projectile is again 1/2m_{vf}^{2} = 1/2(5)(27.92^{2}) = 1948.75 J or rounded to two sig figs, 1900 J