Question

An object is launched with an initial velocity of 10 meters per second from the ground level at an angle of 53° above the horizontal. What is the horizontal range of the object? (cos53°=0,6 sin 53°=0,8).
Take g=10 m/s2

Answer 1

$s=\frac{v^2sin2a}{g}=\frac{2v^2sinacosa}{g}$

$s=\frac{2\cdot10^2\cdot0.6\cdot0.8}{10}=9.6$ (m).