Answer
See the detailed answer below.
Work Step by Step
The given equation is showing that we got a heavy object of 1500 kg that was moving initially at some speed of $v_i$ when it was on the ground and at a height of 10 m its speed became 5 m/s. The energy is conserved.
a) A 1500-kg truck was moving at a speed of $v_i$ when it faces a 10-m height hill. Its speed when it reaches the top of the hill was 5 m/s. If the truck was rolling without friction, find $v_i$.
b) See the graph below.
c)
$$v_{i}=\sqrt{\dfrac{(\frac{1}{2}\times 1500\times 5^2)+(1500\times 9.8\times 10)}{\frac{1}{2}\times 1500}}=\color{red}{\bf 14.9}\;\rm m/s$$
