Answer to Question #314198 in Physics for ABCD

Question

A long jumper leaves the ground at an angle of 20.0° to the horizontal and at a speed of 11.0 m/s.

a. How long does it take for him to reach maximum height? b. What is the maximum height?

c. How far does he jump? (Assume his motion is equivalent to that of a particle, disregarding the motion of his arms and legs.)

A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer makes a grab for the mug and misses, and the mug sails off the end of the bar.

a. How far away from the end of the bar does the mug hit the floor?

b. What are the speed and direction of the mug at impact?

A boy on top of a hill thrown a stone with an initial velocity of 100ft/sec at an inclination of 60° with the horizontal. If the stone has travelled a distance on 400ft from the base of the hill, how high is the hill?

Answer 1

1.

a. How long does it take for him to reach maximum height?

$$t=\frac{v_0\sin\theta}{g}=\frac{11.0\times \sin 20.0^\circ}{9.8}=1.02\;\rm s$$

b. What is the maximum height?

$$h_{\max}=\frac{v_0^2\sin^2\theta}{2g}=\frac{11.0^2\times \sin^2 20.0^\circ}{2\times9.8}=5.15\;\rm m$$

c. How far does he jump?

$$L_{\max}=\frac{v_0^2\sin2\theta}{g}=\frac{11.0^2\times \sin2\times 20.0^\circ}{9.8}=9.20\;\rm m$$

2.

a. How far away from the end of the bar does the mug hit the floor?

$$L_{\max}=v_0\sqrt{2H/g}\\ =1.50\times\sqrt{2\times 1.20/9.81}=0.742\:\rm m$$

b.  What are the speed and direction of the mug at impact?

$$v=\sqrt{v_0^2+2gH}\\ =\sqrt{1.50^2+2\times 9.8\times 1.20}=5.08\rm\: m/s$$

$$\theta=\tan^{-1}\frac{v_y}{v_x}=\tan^{-1}\frac{\sqrt{2gH}}{v_0}$$

$$\theta=\tan^{-1}\frac{\sqrt{2\times 9.81\times 1.20}}{1.50}=72.8^\circ$$