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Chapter 3: Problem 73

A soccer player can kick the ball 28 m on level ground, with its initial velocity at \(40^{\circ}\) to the horizontal. At the same initial speed and angle to the horizontal, what horizontal distance can the player kick the ball on a \(15^{\circ}\) upward slope?

Short Answer

Expert verified
The same player would be able to kick the ball a horizontal distance of D meters (calculate this value using step 3) on a \(15^{\circ}\) upward slope.

Step by step solution

01

Calculate initial Speed

On level ground, maximum horizontal distance (R) formula for a projectile is given by \( R = \frac{{v^2 \sin (2\theta)}}{{g}} \) where \(v\) is initial speed, \(\theta\) is launch angle, and \(g\) is acceleration due to gravity. Given R=28 m, \(\theta=40^{\circ}\) and \(g=9.81 m/s^2\), we can calculate initial speed \(v\) by rearranging the formula: \( v = \sqrt{\frac{{R\cdot g}}{{\sin (2\theta)}}} \)
02

Calculate new Launch Angle

The slope adds an angle of \(15^{\circ}\) to the original projection angle. Therefore, the new launch angle \( \alpha = \theta + 15^{\circ} \)
03

Calculate Horizontal Distance on Slope

Now, we use the new launch angle and the initial speed calculated in step 1 into the range D formula on sloping ground: \( D = \frac{2v^2 \sin (\alpha) \cos (\alpha - 15^{\circ})}{g} \)

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