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
40m∕ s and the launch angle is
37∘, we can use the sine function to find the vertical component of the initial velocity
(V0y) as follows:
V0y=V0⋅sin‌(θ)Where:
V0y is the vertical component of the initial velocity.
V0=40m∕ s is the initial velocity.
θ=37∘ 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∘,45∘, and
60∘ angles in a right triangle do not directly match the given ratio. However, the
37∘ 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∘ 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∘ (approximately 0.6 based on trigonometric tables):
V0y=40⋅sin‌(37∘)≈40⋅0.6=24m∕ sThe 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
9.8m∕ s2 downward.
t=‌=‌≈2.45 sThus, the time at which the ball reaches the highest point of its path is approximately
2.45 s, which is closest to Option A (2.4 s).