Question #306986

Suppose a golf ball is hit off with the tee with an initial velocity of 30 m/s at an angle of 35° to the horizontal.


What is the maximum height reached by the ball?


What is its range?



1
Expert's answer
2022-03-13T18:50:17-0400

The maximum height is given as follows (see https://courses.lumenlearning.com/boundless-physics/chapter/projectile-motion/):


h=u2sin2θ2gh = \dfrac{u^2\sin^2\theta}{2g}

where u=30m/s,θ=35°,g=9.8m/s2u = 30m/s, \theta = 35\degree , g = 9.8m/s^2. Thus, obtain:


h=302sin35°29.826mh = \dfrac{30^2\cdot \sin35\degree}{2\cdot 9.8} \approx 26m

The range is:


R=u2sin2θg=302sin235°9.886mR = \dfrac{u^2\sin2\theta}{g} = \dfrac{30^2\sin2\cdot 35\degree}{9.8} \approx 86m

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