A ball is projected at an angle of 30° above the horizontal with a speed of 35 m/s. What will be its approximate horizontal range across a level surface assuming no air resistance? F. 17.5 m
G. 54.2 m
H. 108 m
J. 125 m

The horizontal distance the ball travels before returning to the ground is 19.5 m

Step-by-step explanation:

Formula of range of horizontal distance =\frac{u^2 Sin 2 \theta}{g}gu2Sin2θ

Where u is the initial velocity

g is gravity

We are given that A ball is kicked off the ground with a velocity of 15 m/s at an angle of 30° horizontally.

So, u = 15 m/s

\theta = 30^{\circ}θ=30∘

g = 10

So, the horizontal distance the ball travels before returning to the ground = \frac{15^2 Sin 2(30^{\circ})}{10}=19.48 \sim 19.5 m10152Sin2(30∘)=19.48∼19.5m

Hence The horizontal distance the ball travels before returning to the ground is 19.5 m

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