A ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 50.

Question

A ball is thrown upward with an initial speed of 20.0 m/s from the edge of a 50.0 m high cliff. At the in stand the ball is thrown, a woman stars running away from the base of the cliff with a constant speed of 5.00 m/s. The woman runs in a straight line on level ground, and air resistance acting on the ball can be ignored. a. At what angle above the horizontal should the ball be thrown so that runner will catch it just before it hits the ground? b. How far does the woman ran before she catches the ball?

Explanation / Answer

a)let the time of flight be t

and angle be a w.r.t the horizontal

X = Ux*t

5t = 20 cos a *t

cos a = 1/4

a = 75.5o

b) Y = Uyt - 0.5gt^2

-50 = 20 sin 75.52o t - 0.5*9.8*t^2

4.9t^2 - 19.36t -50 =0

t = 19.36 + sqrt(19.36^2 + 4*4.9*50) / 9.8

= 5.73 s

X = 5t = 5*5.73 = 28.65 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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