Question

A rock is launched by a catapult with an initial velocity of 52 feet per second and an angle of 57∘ to the horizontal. if the rock is launched from an initial height of 6 feet, how far from the launch point does the rock land?

Answer 1

To find the horizontal displacement of the rock, you can use the equations Vx = V * cos(theta) and d = Vx * t, where V is the initial velocity, theta is the launch angle, and t is the time of flight. The horizontal component of velocity can be found using Vx = 52 * cos(57). The time of flight can be found using t = 2 * (52 * sin(57)) / 32.2. Finally, the horizontal displacement can be found using d = (52 * cos(57)) * (2 * (52 * sin(57)) / 32.2).

Step-by-step explanation:

To find the horizontal displacement of the rock, we can break down the initial velocity into horizontal and vertical components. The horizontal component can be found using the equation Vx = V * cos(theta), where V is the initial velocity and theta is the launch angle. Plugging in the given values, we have Vx = 52 * cos(57). The vertical component can be found using Vy = V * sin(theta), where V is the initial velocity and theta is the launch angle. Plugging in the given values, we have Vy = 52 * sin(57).

Next, we can find the time it takes for the rock to hit the ground. The equation for the time of flight is t = 2 * Vy / g, where Vy is the vertical component of velocity and g is the acceleration due to gravity. Plugging in the given values, we have t = 2 * (52 * sin(57)) / 32.2.

Finally, the horizontal displacement can be found using the equation d = Vx * t, where Vx is the horizontal component of velocity and t is the time of flight. Plugging in the values we calculated earlier, we have d = (52 * cos(57)) * (2 * (52 * sin(57)) / 32.2).