Answer to Question #287040 in Physics for omar

Question

 A golf ball is launched from the roof of a school with a velocity of 20 m/s at an angle of 300 above the horizontal. If the roof is 40 m above the ground, calculate:

a) the ball’s time of flight

b) the ball’s horizontal displacement

Answer 1

a) The time of flight:

$$t=\dfrac{v\sin\theta}{g}+\sqrt{\dfrac{2(h+((v\sin\theta)^2/(2g))}{g}}=4.05\text{ s}.$$

b) The range:

$$R=v\cos\theta·t=70.2\text{ m}.$$