Question

Joe stands 9.50 m from the door. Joe throws the ball toward the door giving an initial velocity of 11.9 m/sec at an angle of 28.0 degrees above the horizontal. joe released the ball 1.20 m above the ground .
A) how long after joe releases the bal lwill it hit the door?
b) at what height above the ground will the ball hit the door?.

Answer 1

the equation for the motion of a body in horizontal direction is : \[x = u \cos \theta \times t\]
here, x and ucostheta are known quantities. so find "t".
after having found the value of "t", write down the equation for motion of body in vertical direction and that should give yo the height at which the ball would strike the door.

Answer 2

As shown above you can find the time for which ball will travel the distance to the door
Which is 0.90 s
Then using kinematics equation
\[h = h _{0}+v _{0y} \times t - (g t ^{2})/2\]
where
h0, t are already known and
\[v _{0y}=v \sin 28^{0}=5.59 m/s\]
using this you will find that
h - the height above the ground at which the ball will hit the door is equal to
2.18 m assuming that g=10m/s2