Question

(1) A stone is thrown obliquely upwards with the output speed 25m / s from a point 1.5m above ground level. The throwing height of the bow is 26.0m above the ground. how big is the stone's speed when it passes the highest point? we ignore the air resistance.

Answer 1

The law of conservation of the mechanical energy gives

$$\frac{mv_i^2}{2}+mgh_i=\frac{mv_f^2}{2}+mgh_f$$

So, final speed

$$v_f=\sqrt{v_i^2-2g(h_f-h_i)}$$

$$=\sqrt{(25\:\rm{m/s})^2-2\times 9.8\:\rm{m/s^2}\times (26\:\rm{m}-1.5\:\rm{m})}$$

$$=12\:\rm{m/s}$$