Question

A golf ball is hit with a speed of 55 m/s at an angle of 40° to the horizontal. Neglecting air resistance (a) (5 pts.) What are the x- and y-components of the initial velocity? (b) (5 pts.) To what height does the ball rise? (c) (5 pts.) What are the magnitude and direction of the ball's velocity at the maximum height? (d) (5 pts.) What is the total time in the air? (e) (5 pts.) How far does the ball travel horizontally? (1) (5 pts.) For the same initial speed, what other initial angle of projection would result in the same horizontal range?

Answer 1

B veose Gunen initial speed =55m/s profection angle d = 40 AR R ② X-componant of initial velocity Vx = v case -55 cos(40) V = 42-132 m/s y component of initial velocity Vy= usine = 55 Sin(40) uy = 35.353 mis height h = 62sino zg - (55² sinuos - 3025x0.41317 = 63.76 2x9.0 h = 63.76 m A 19.6 & At maximum height (ire point B) Only x-componant of initial velocity will be there. Hence at maximum height Velocity 16 = Vx = 42 -132 m/s Aus O angle o=0 (from horizontal ) tolal time in the air (or line of flyte) @ T= 2usino = 2x55x Sin(40) = 0. 7066 og god 9.0 T = 7.21 see an Horizontal distance covered by the ball (or forange ) is given by R=vSinzo = 55) sin(2x40) - 3025% 0.9898 @ ruired angle o=90 0 303.98 mida F 90-40 = ²50 ch