Question

Boat A moves with constant velocity of 6m/s. starting from position down. Fine the angle in order for the projectile to hit the boat 5 seconds after starting, under the conditions given. How high is the hill above the water?

Answer 1

$vt+90=30\cdot\cos\alpha\cdot t\to \cos\alpha=\frac{vt+90}{30t}=\frac{6\cdot5+90}{30\cdot5}=0.8\to$

$\alpha=36.87°$ . Answer

$t_1=\frac{30\cdot\sin36.87°}{9.8}=1.84\ (s)$

The time of movement of the projectile in the vertical direction

$t_2=5-1.84=3.16\ (s)$

$h=gt^2/2=9.8\cdot3.16^2/2\approx49\ (m)$

$h_1=(30\cdot\sin36.87°)^2/(2\cdot9.8)=16.53\ (m)$

$y=49-16.53=32.47\ (m)$ . Answer