A golf ball is struck by a club such that it leaves the ground with an initial velocity of $40 \text{ m s}^{-1}$ at an angle of $60^{\circ}$ with the horizontal. Considering air resistance to be negligible, what is the maximum vertical height reached by the ball?
(Take $g = 10 \text{ m s}^{-2}$ )

Question

A golf ball is struck by a club such that it leaves the ground with an initial velocity of

40 \text{ m s}^{-1}

at an angle of

60^{\circ}

with the horizontal. Considering air resistance to be negligible, what is the maximum vertical height reached by the ball?
(Take

g = 10 \text{ m s}^{-2}

)

Quantrex Academy · Accepted Answer

Correct answer: B. 60 m

Initial velocity, u = 40 m s^ -1 Angle of projection, θ = 60^ Acceleration due to gravity, g = 10 m s^ -2 The maximum vertical height H is given by the formula: H = (u^2 ^2 θ)/(2g) Substituting the given values: H = (40)^2 ^2 60^ 2 × 10 H = 1600 × ( 3 2 )^2 20 H = (1600 × (3)/(4))/(20) H = (1200)/(20) = 60 m Answer: 60 m

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