Question: 37: A ball is projected with a velocity, 10 \mathrm{~ms}^{-1}, at an angle of 60^{\circ} with the vertical direction. Its speed at the highest point of its trajectory will be
(1) 5 \mathrm{~ms}^{-1}
(2) 10 \mathrm{~ms}^{-1}
(3) Zero
(4) 5 \sqrt{3} \mathrm{~ms}^{-1}
Answer: Option (4)
Explanation:
The initial velocity of the ball is u = 10 \mathrm{~ms}^{-1},
and it is projected at an angle of 60^{\circ} with the vertical.
To find the speed at the highest point, we note that the vertical component of velocity becomes zero at the highest point, while the horizontal component remains unchanged.
The horizontal component of velocity is given by
u_h = u \sin 60^{\circ}, because the angle is measured with the vertical.
Substituting the values, u_h = 10 \times \sin 60^{\circ}.
Since \sin 60^{\circ} = \frac{\sqrt{3}}{2},
we get u_h = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \mathrm{~ms}^{-1}.
At the highest point, the speed of the ball is equal to its horizontal component of velocity.
Therefore, the speed at the highest point is 5 \sqrt{3} \mathrm{~ms}^{-1}.
Hence, the correct answer is Option (4).