Question 28: Laws of Motion

Question

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

Accepted Answer

To find the time at which the ball reaches the highest point of its path, it's essential to analyze the motion in vertical direction, taking into account that only the vertical component of the initial velocity affects the time to reach the highest point. We can decompose the initial velocity into its horizontal and vertical components using trigonometric functions.
Given that the initial speed is $$40 \, 	ext{m/s}$$ and the launch angle is $$37^{\circ}$$, we can use the sine function to find the vertical component of the initial velocity ($$V_{0y}$$) as follows:
$$V_{0y} = V_0 \cdot \sin(	heta)$$
Where:
$$V_{0y}$$ is the vertical component of the initial velocity.
$$V_0 = 40 \, 	ext{m/s}$$ is the initial velocity.
$$	heta = 37^{\circ}$$ is the angle of projection.
From the given triangle ratio $$6:8:10$$, which corresponds to the ratio of the sides in a right triangle, we recognize the pattern of a $$3:4:5$$ triangle scaled up. This triangle ratio is typical for illustrating trigonometric relationships, where the sides opposite to the $$30^{\circ}$$, $$45^{\circ}$$, and $$60^{\circ}$$ angles in a right triangle do not directly match the given ratio. However, the $$37^{\circ}$$ angle here is used to show an example, not derived from the triangle's side lengths. Since the specific ratio does not correspond to the $$37^{\circ}$$ angle in a straightforward trigonometric way, but rather serves as a background for the problem context involving trigonometric calculation, let's proceed with the calculation for the vertical component of velocity.
Using the sine of $$37^{\circ}$$ (approximately $$0.6$$ based on trigonometric tables):
$$V_{0y} = 40 \cdot \sin(37^{\circ}) \approx 40 \cdot 0.6 = 24 \, 	ext{m/s}$$
The time to reach the highest point can be found by dividing the vertical component of the initial velocity by the acceleration due to gravity (assuming the only force acting on the ball is gravity), which decelerates the ball until it momentarily stops at the peak of its trajectory. The acceleration due to gravity ($$g$$) is $$9.8 \, 	ext{m/s}^2$$ downward.
$$t = \frac{V_{0y}}{g} = \frac{24 \, 	ext{m/s}}{9.8 \, 	ext{m/s}^2} \approx 2.45 \, 	ext{s}$$
Thus, the time at which the ball reaches the highest point of its path is approximately $$2.45 \, 	ext{s}$$, which is closest to Option A (2.4 s).

Answer

2.4 s

Answer

0.32 s

Answer

3.2 s

Answer

0.24 s

Laws of Motion

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