How high is the wall if a dropped object hits ground in 1.1 seconds?
h = (1/2) g t^2
= 4.9(1.1)^2 = 5.93 meters high
so
u = 54 cos 34 forever
Vi =54 sin 34
v = 54 sin 34 - 9.8 t
h = 5.93 + Vi t - 4.9 t^2
when h = 0, the ball hits ground
0 = 5.93 + (54 sin 34) t - 4.9 t^2
solve for t with quadratic eqn
then
distance = u t = 54 cos 34 * t
for the last part, v = 0 at the top so
9.8 t = 54 sin 34
use that t in
h = 5.93 + (54 sin 34) t - 4.9 t^2
to get h at the top
King Arthur's knights fire a cannon from the top of the castle wall. The cannonball is fired at a speed of 54 m/s and at an angle of 34°. A cannonball that was accidentally dropped hits the moat below in 1.1 s.(a) (a)How far from the castle wall does the cannonball hit the ground?
(b) What is the ball's maximum height above the ground?
HELP!
1 answer
2 answers