Question

A projectile is fired from ground level at time t = 0, at an angle \theta with respect to the horizontal. It has an initial speed v_{0}. In this problem, we are assuming that the ground is level.

Part A. Find the time t_{H} it takes the projectile to reach its maximum height. Express t_{H} in terms of v_{0}, \theta, and g (the magnitude of the acceleration due to gravity).

Part B. Find t_{R}, the time at which the projectile hits the ground. Express the time in terms of v_{0}, \theta, and g.

Part C. Find H, the maximum height attained by the projectile. Express the maximum height in terms of v_{0}, \theta, and g.

Part D. Find the total distance R (often called the range) traveled in the x-direction; in other words, find where the projectile lands. Express the range in terms of v_{o}, \theta, and g.

Answer 1

To find the time it takes for the projectile to reach its maximum height, use the equation t_H = v_0 sin(θ) / g. To find the time the projectile hits the ground, use the equation t_R = 2v_0 sin(θ) / g. The maximum height attained by the projectile can be found using the equation H = (v_0^2 sin^2(θ)) / (2g). The total distance traveled in the x-direction (range) can be found using the equation R = (v_0^2 sin(2θ)) / g.

Step-by-step explanation:

To find the time it takes for the projectile to reach its maximum height, we can use the equation for vertical motion:

t_H = v_0 sin(\theta) / g

To find the time the projectile hits the ground, we can use the equation for vertical motion:

t_R = 2v_0 sin(\theta) / g

The maximum height attained by the projectile can be found using the equation:

H = (v_0^2 sin^2(\theta)) / (2g)

The total distance traveled in the x-direction (range) can be found using the equation:

R = (v_0^2 sin(2\theta)) / g

Answer 2

(A) t_H = Vo×Sin (theta)/g

(B) t,_R = (2VoSin(theta))/g

(C) H = Vo²Sin²(theta) / 2g

(D) Range = Vo²Sin(2×theta) / g.

Step-by-step explanation:

The detailed solution to the problem can be found in the attachment below