## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 4 - Problems - Page 150: 45

#### Answer

The skater should start this stunt a distance of 15.8 meters from the bottom of the ramp.

#### Work Step by Step

We can find the time it takes to fall 3.00 meters: $\Delta y = \frac{1}{2}a_y~t^2$ $t = \sqrt{\frac{2\Delta y}{a_y}}$ $t = \sqrt{\frac{(2)(3.00~m)}{9.80~m/s^2}}$ $t = 0.7825~s$ We can find the required horizontal velocity to travel 7.00 meters in this time: $v_x = \frac{7.00~m}{0.7825~s} = 8.946~m/s$ We can find the acceleration while moving down the ramp: $mg~sin~\theta = ma$ $a = g~sin~\theta$ $a = (9.80~m/s^2)~sin~15.0^{\circ}$ $a = 2.536~m/s^2$ We can find the distance the skater needs to travel down the ramp in order to have a velocity of $8.946~m/s$ at the bottom: $v_f^2 = v_0^2+2ad$ $d = \frac{v_f^2-v_0^2}{2a}$ $d = \frac{(8.946~m/s)^2-0}{(2)(2.536~m/s^2)}$ $d = 15.8~m$ The skater should start this stunt a distance of 15.8 meters from the bottom of the ramp.

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