A ball is shot from the ground into the air. At a height of 9.1 m, the velocity

Question

A ball is shot from the ground into the air. At a height of 9.1 m, the velocity is observed to be v = (7.5 + 6.2 ) 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

Explanation / Answer

at height 9.1m v = (7.5 + 6.2) m/s horizontal and upward

Horizontal velocity is constant.

for upward
v^2 - u^2 = 2as

7.5*7.5 - u^2 = 2(-9.8)9.1
u^2 = 234.61
u = 15.32 m/s initial velocity in y direction

Initial speed = (6.22 + 15.322) = 16.53m/s

tan = 15.32/6.2

= 67.96 degrees

a) Maximum height = u^2/2g = 15.32*15.32/(2*9.8) = 11.95 m

where u is initial velocity in y direction

b) horizontal distance = 16.53*16.53 sin2 /g

                                 = 16.53*16.53 sin(2*67.96) / 9.8 = 19.4 m

c)Magnitude of ball velocity just before it hits ground = magnitude of ball velocity at (t=0)

= 16.53m/s

d) = 67.96degrees

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.