Q166 / 180 · COMEDK 2023 (Evening Shift) · Physics
A hockey player hits the ball at an angle of $37^{\circ}$ from the horizontal with an initial speed of $40 \mathrm{~m} / \mathrm{s}$ (a right angled triangle with one of the angle is $37^{\circ}$ and their sides in the ratio of $6: 8: 10$). Assume that the ball is in a vertical plane. The time at which the ball reaches the highest point of its path is
- A3.2 s
- Correct answerB2.4 s
- C0.32 s
- D0.24 s
Solution
The initial velocity of the ball is $u = 40 \text{ m/s}$ and the angle of projection is $\theta = 37^{\circ}$.
The vertical component of the initial velocity is given by $u_y = u \sin \theta$.
Given the triangle with sides in ratio $3:4:5$ (which is equivalent to $6:8:10$), $\sin 37^{\circ} = \dfrac{3}{5}$.
Substituting the values, $u_y = 40 \times \dfrac{3}{5} = 8 \times 3 = 24 \text{ m/s}$.
At the highest point of the trajectory, the vertical component of the velocity becomes zero, $v_y = 0$.
Using the equation of motion $v_y = u_y - gt$, where $g = 10 \text{ m/s}^2$ is the acceleration due to gravity:
$0 = 24 - 10t$
$10t = 24$
$t = 2.4 \text{ s}$.
Answer: 2.4 s
The vertical component of the initial velocity is given by $u_y = u \sin \theta$.
Given the triangle with sides in ratio $3:4:5$ (which is equivalent to $6:8:10$), $\sin 37^{\circ} = \dfrac{3}{5}$.
Substituting the values, $u_y = 40 \times \dfrac{3}{5} = 8 \times 3 = 24 \text{ m/s}$.
At the highest point of the trajectory, the vertical component of the velocity becomes zero, $v_y = 0$.
Using the equation of motion $v_y = u_y - gt$, where $g = 10 \text{ m/s}^2$ is the acceleration due to gravity:
$0 = 24 - 10t$
$10t = 24$
$t = 2.4 \text{ s}$.
Answer: 2.4 s
Hence, the correct answer is option: 'B'.