A projectile is fired from the ground, reaches a maximum height of 70.7 m and lands a distance of 29.2 m away from the launch point. What was the projectile s launch velocity? 63.6 m/s, 12.6 degrees above horizontal 37.4 m/s, 84.1 degrees above horizontal 37.2 m/s, 42.1 degrees above horizontal 3.8 m/s, 25.2 degrees above horizontal

Name: A projectile is fired from the ground, reaches a maximum height of 70.
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Description: A projectile is fired from the ground, reaches a maximum height of 70.7 m and lands a distance of 29.2 m away from the launch point. What was the projectile s launch velocity? 63.6 m/s, 12.6 degrees above horizontal 37.4 m/s, 84.1 degrees above hor

we have given, Maximum height (Hmax) = 70.7 m Hmax is given by Hmax = u2sin2θ2g = 70.7 u2sin2θ =…

Answered: A projectile is fired from the ground,…

College Physics

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A projectile is fired from the ground, reaches a maximum height of 70.7 m and lands a distance of 29.2 m away from the launch point. What was the projectile s launch velocity?

		63.6 m/s, 12.6 degrees above horizontal
		37.4 m/s, 84.1 degrees above horizontal
		37.2 m/s, 42.1 degrees above horizontal
		3.8 m/s, 25.2 degrees above horizontal

Expert Solution

Step 1

we have given,

Maximum height (H_max) = 70.7 m

H_max is given by

H_max = $\frac{u^{2} \sin^{2} θ}{2 g}$ = 70.7

$u^{2} \sin^{2} θ = 70.7 x 2 g = 1385.72$ - - - -- -- - - - (i)

Horizontal distance or Range (R) = 29.2 m

Range R is given as,

R = $u \times \cos θ \times T = \frac{u \times \cos θ \times 2 \times u \times \sin θ}{g} = \frac{2 u^{2} \sin θ \cos θ}{g} = 29.2$ - - - - - (ii)

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