A projectile is released at a velocity of 15 m/s at an angle of 15deg above the horizontal. At what height should the projectile be released to hit a target 15 m away at ground level? (Ignore air resistance. Assume uniform acceleration due to gravity = -9.8 m/s2)

Question

anonymous · Answer

use the formula
t(sec)=x/(vocos15)=15/(15 cos 15)=1/cos 15= 1.0353
height H=(vo sin15)t - 0.5gt^2= (15 sin 15)1.0353-0.5(9.8)(1.0353)^2= -1.233 meter