Answer:
(a) 1.73 s
(b) 14.75 m
(c) 3.36 s
(d) double
(e) 63.32 m
Explanation:
Vertical component of initial velocity, uy = 17 m/s
Horizontal component of initial velocity, ux = 18.3 m/s
(A) At highest point of trajectory, the vertical component of velocity is zero. Let the time taken is t.
Use first equation of motion in vertical direction
vy = uy - gt
0 = 17 - 9.8 t
t = 1.73 seconds
(B) Let the highest point is at height h.
Use III equation of motion in vertical direction
0 = 17 x 17 - 2 x 9.8 x h
h = 14.75 m
(C) The time taken by the ball to return to original level is T.
Use second equation of motion i vertical direction.
h = 0 , u = 17 m/s
0 = 17 t - 0.5 x 9.8 t^2
t = 3.46 second
(D) It is the double of time calculated in part A
(E) Horizontal distance = horizontal velocity x total time
d = 18.3 x 3.46 = 63.32 m
