Answered


7. A racquetball thrown from the ground at an angle of 45° and with a speed of 22.5 m/s
lands exactly 2.5 s later on the top of a nearby building. Calculate the horizontal distance it
traveled and the height of the building.​

Sagot :

C41992viz

Answer:

HD = 39.77m

Height of building = 9.131m

Explanation:

s = Vyt - 0.5gt²

= (22.5sin(45))(2.5) - 0.5(9.806)(2.5)²

= 9.131m (apprx)

x = Vxt

= (22.5 cos45)(2.5)

= 39.77 (apprx)