Question

WUJ 30' with
19. A cannon shell fired from the edge of a cliff with a velocity of 100 m/s in a direction 3
the horizontal, strikes the level ground at the bottom of the cliff 1 km from the base. How
does it take for the shell to hit the ground and how high is the cliff?
(11.55 s; 76:​

Answer 1

Answer:

$(100 \cos(30)) t = 1000$

$t = 20 \div \sqrt{3} = 11.55$

$h = (50)(11.55) - \frac{1}{2} (10)(11.55) {}^{2} = 76$