14 A sportsman hits a ball with a bat.
The ball leaves the bat at a speed of $37 \mathrm{~ms}^{-1}$ at an angle of $53^{\circ}$ to the horizontal, as shown.

Calculate the horizontal distance travelled by the ball before returning to the height it was hit from.

Question

14 A sportsman hits a ball with a bat.
The ball leaves the bat at a speed of $37 \mathrm{~ms}^{-1}$ at an angle of $53^{\circ}$ to the horizontal, as shown.

Calculate the horizontal distance travelled by the ball before returning to the height it was hit from.

Accepted Answer

∻Solution∻ ‖ a ⋮ First, we need to resolve the initial velocity into its horizontal and vertical components. The horizontal component is given by: $$v_{0x} = v_0 \cos( heta) = 37 \cos(53^\circ)$$. Calculating this gives us: $$v_{0x} \approx 22.2 \, \mathrm{m/s}$$. ‖ ‖ b ⋮ The vertical component of the initial velocity is: $$v_{0y} = v_0 \sin( heta) = 37 \sin(53^\circ)$$. This results in: $$v_{0y} \approx 29.6 \, \mathrm{m/s}$$. ‖ ‖ c ⋮ To find the time of flight, we can use the formula for the vertical motion. The time to reach the maximum height is given by: $$t_{up} = \frac{v_{0y}}{g} = \frac{29.6}{9.8} \approx 3.02 \, \mathrm{s}$$. The total time of flight is then: $$T = 2t_{up} \approx 6.04 \, \mathrm{s}$$. ‖ ‖ d ⋮ The horizontal distance traveled (range) can be calculated using: $$d = v_{0x} \cdot T = 22.2 \cdot 6.04 \approx 134.9 \, \mathrm{m}$$. ‖ ∻Answer∻ ⚹ 134.9 m ⚹ ∻Key Concept∻ ⚹ Projectile Motion: The motion of an object thrown into the air, subject to the acceleration due to gravity. The horizontal and vertical motions are independent of each other. Key equations include: $$d = v_{0x} \cdot T$$ for horizontal distance and $$T = 2 \frac{v_{0y}}{g}$$ for total time of flight. ⚹ ∻Explanation∻ ⚹ By resolving the initial velocity into horizontal and vertical components and using the equations of motion, we can determine the total horizontal distance traveled by the ball before it returns to its original height. ⚹

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