A ball is thrown at a speed of 40 m/s at an angle of 60° with the horizontal. Find (a) the maximum height reached and (b) the range of the ball. Take g = 10 m/s².

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Solution

Given:
Initial speed of the ball, u = 40 m/s
Angle of projection of the ball with the horizontal, α = 60°
Also,
a = g = 10 m/s²

(a) Maximum height reached by the ball:
$H = \frac{u^{2} \sin^{2} α}{2 g}$
$\Rightarrow H = \frac{40^{2} {(\sin 60 °)}^{2}}{2 \times 10} = 60 m$

(b) Horizontal range of the ball:
$R = \frac{u^{2} \sin 2 α}{g}$
$= \frac{40^{2} \sin (2 \times 60 °)}{10} = 80 \sqrt{3} m$

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A ball is thrown at a speed of 40 m/s at an angle of 60° with the horizontal. Find (a) the maximum height reached and (b) the range of the ball. Take g = 10 m/s2.

Given: Initial speed of the ball, u = 40 m/s Angle of projection of the ball with the horizontal, α = 60° Also, a = g = 10 m/s2 (a) Maximum height reached by the ball: H=u2sin2α2g ⇒H=402 sin 60°22×10=60 m (b) Horizontal range of the ball: R=u2sin2αg =402 sin2×60°10=803 m

A ball is thrown at a speed of 40 m/s at an angle of 60° with the horizontal. Find (a) the maximum height reached and (b) the range of the ball. Take g = 10 m/s².