Answer to Question #158711 in Field Theory for Arslan habib

Question

A person is riding on a flatcar traveling at a constant speed of 200 km/s (Fig.). He wishes to throw

a ball through a stationary hoop 90 km above the height of his hands in such a manner that the ball

will move horizontally as it passes through the hoop. He throws the ball with a speed of 250 m/s

with

respect to himself,

(a) How many time required after he releases the ball will it pass through the hoop?

(b) What must the vertical component of the initial velocity of the ball?

(c) At what horizontal distance in front of the hoop must he release the ball?

(d) When the ball leaves the man's hands, what is the direction of its velocity relative to the

frame of reference of the flatcar? Relative to the frame of reference of an observer standing

on the ground?

90 km

V

V

X

v = 200 km/s

Answer 1

b)

$$v_{0y}^2=2gh\\v_{0y}^2=2(9.8)(90)\\v_{0y}=42\frac{m}{s}$$

a)

$$v_{0y}=gt\\42=9.8t\\t=4.3\ s$$

c)

$$v_{0x}=\sqrt{250^2-42^2}=246\frac{m}{s}$$

$$(246+200)4.3=x\\x=1920\ m$$

d)

$$\alpha=\arctan{\frac{42}{246}}=9.7\degree\\ \beta=\arctan{\frac{42}{246+200}}=5.4\degree$$