A stone is thrown with an initial speed of 14.8 m/s at an angle of 51° above the horizontal from the top of a 35 m building.If g = 9.8 m/s² and air resistance is negligible,then what is the magnitude of the vertical velocity component of the rock as it hits the ground?

Question

Quizplus · Accepted Answer

First, we can use the initial velocity to find the vertical and horizontal components of the velocity:

Vertical component: V_y = V_i*sin(theta) = 14.8*sin(51) = 11.5 m/s

Horizontal component: V_x = V_i*cos(theta) = 14.8*cos(51) = 10.8 m/s

Next, we can use the vertical component of the velocity to find the time it takes for the stone to hit the ground:

y = V_i*t + (1/2)*g*t^2

35 = 14.8*sin(51)*t + (1/2)*9.8*t^2

Solving for t, we get t = 3.77 s

Finally, we can use the time and horizontal component of the velocity to find the distance the stone travels before hitting the ground:

x = V_x*t = 10.8*3.77 = 40.8 m

Therefore, the magnitude of the velocity as the stone hits the ground is:

V = sqrt((V_x)^2 + (V_y - g*t)^2) = sqrt((10.8)^2 + (11.5 - 9.8*3.77)^2) = 29 m/s

Therefore, the answer is D.

A Stone Is Thrown with an Initial Speed of 14