Question

A hockey player hits the ball at an angle of 37° from the horizontal with an initial speed of 40 m/s (a right-angled triangle with one of the angles being 37° 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:

• A. 2.4 s
• B. 0.32 s
• C. 3.2 s
• D. 0.24 s

Hint:

To calculate the time to reach the highest point, use the formula \( t = \frac{u_y}{g} \), where \( u_y \) is the vertical component of the initial velocity, and \( g \) is the acceleration due to gravity.

Answer 1

The Correct Option is A

We are given the initial speed of the ball as \( u = 40 \, \text{m/s} \), and the angle of projection \( \theta = 37^\circ \). We need to find the time at which the ball reaches the highest point of its path. The initial velocity can be broken into two components: - The horizontal component of velocity, \( u_x = u \cdot \cos(\theta) \), - The vertical component of velocity, \( u_y = u \cdot \sin(\theta) \). Now, substituting the values: \[ u_x = 40 \cdot \cos(37^\circ) \quad \text{and} \quad u_y = 40 \cdot \sin(37^\circ) \] Using \( \cos(37^\circ) \approx 0.7986 \) and \( \sin(37^\circ) \approx 0.6018 \), we get: \[ u_x \approx 40 \cdot 0.7986 = 31.944 \, \text{m/s} \quad \text{and} \quad u_y \approx 40 \cdot 0.6018 = 24.072 \, \text{m/s} \] At the highest point of the trajectory, the vertical velocity becomes zero. The time \( t \) to reach the highest point is given by the equation: \[ v_y = u_y - g \cdot t \] Where: - \( v_y \) is the final vertical velocity (which is zero at the highest point), - \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity, - \( t \) is the time to reach the highest point. At the highest point: \[ 0 = 24.072 - 9.8 \cdot t \] Solving for \( t \): \[ t = \frac{24.072}{9.8} \approx 2.45 \, \text{seconds} \] Thus, the time taken to reach the highest point of the path is approximately 2.4 seconds. Therefore, the correct answer is (A) 2.4 s.