A ball is shot from the ground into the air. At a height of 9.1 m, the velocity is observed to be v = (6.6 + 5.7 ) m/s, with horizontal and upward.

(a) To what maximum height does the ball rise?
m

(b) What total horizontal distance does the ball travel?
m

(c) What is the magnitude of the ball's velocity just before it hits the gound?
m/s

(d)What is the direction of the ball's velocity just before it hits the ground?
° relative to the horizontal

3 answers

6.7²/(2g) + 9.1 = 11.39 m

b) t = 2√[2H/g] = 3.05 sec; X = 6.6*t = 20.125 m

c) Vf = (6.6i - 6.7j) m/s

d) Θ = arctan[j/i] = arctan[-6.7/6.6] = -45.43°

mali yan

Yall wrong