Question

A basketball was kicked 20.0 m above horizontal with an initial velocity of 120.0 m/s at an angle of 35.0˚ above the horizontal. (a) How long is the rocket in air before falling? (b) What is the horizontal range for the rocket? (c) With what speed does the rocket hit the ground? Please show work

Answer 1

A ball is projected at an angle Swith horizontal with an initial velocity (u).Resolve vel ocity into components - (120 m/s)cos 35* 98.298m/s -(120 m/s) sin 35* 68.8 m/s The time spent(time of flight) rocket in air can be calculated as folows gt 2u t=- 2(68.8 m/s) 9.8m/s2 = 14.05s~ 14 s Range of the projectile is, 2u,i, 2(98.298m/s) (68.8 m/s) 9.8 m/s2 - 1380 m or 1.380 km s1.4km During the ball reaches the plane of projection Horizontal velocity u, ucos -98.298m/s Vertical velocity (688 m/s)+2(98m/s)(20m) 71.59m/s Resultant velocity is (98.298m/s(71.59m/s) 121.6m/s