Harry is on the 10h floor (approximately 100 feet above ground level), and throw

Question

Harry is on the 10h floor (approximately 100 feet above ground level), and throws a ball upward with initial speed of 160 feet/sec. Note: positive Xis upward, and g-32 feet/sec1 Hint:x) and v) vo + at (a) What is the maximum height above the ground the ball reaches? (b) How long does it take to reach this maximum height? (c) When the ball comes back down and passes Harry (at 100 feet elevation), what is its velocity? (d) At what speed will the ball hit the ground? (e) At what time will the ball hit the ground? (f) What is the acceleration of the ball at the maximum height? (g) What is the acceleration of the ball just before it hits the ground?

Explanation / Answer

Given that,

yo = 100 ft

u = 160 ft/s

(a)

At maximum height,

v = 0

From kinematic equation,

v^2 - u^2 = 2as

0 - (160)^2 = 2(-32)*h

h = 400 ft

Maximum height above ground,

H = 100 + 400

H = 500 ft

(b)

At maximum height, v = 0

v = u + at

0 = 160 - 32*t

t = 5 s

(c)

Whenthe ball comes back down and passes harry,

velocity, v = initial velocity

v = 160 ft/s

(d)

v^2 - u^2 = 2as

Here u = 0 (at maximum height)

v = speed with which ball hits the ground.

v^2 - 0 = 2*32*500

v = 178.88 m/s

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

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