A stone is thrown with an initial speed of 10.3 m/s at an angle of 51° above the horizontal from the top of a 32 m building.If g = 9.8 m/s² and air resistance is negligible,then what is the speed of the rock as it hits the ground?

Question

Quizplus · Accepted Answer

Using the kinematic equations, we can find the time it takes for the rock to hit the ground. 
Firstly, we find the vertical component of the initial velocity: 
Viy = 10.3 m/s * sin(51°) = 7.87 m/s

Next, let's use the kinematic equation: 
y = Viy*t + 1/2 * a * t^2 where y = -32 m, Viy = 7.87 m/s, and a = -9.8 m/s^2

-32 m = (7.87 m/s)t + 1/2 * (-9.8 m/s^2) * t^2

Solving for t, we get t = 3.23 s

Now that we know the time it takes for the rock to hit the ground, we can find the horizontal distance it travels:

Vix = 10.3 m/s * cos(51°) = 6.60 m/s

x = Vix * t = 6.60 m/s * 3.23 s = 21.3 m

The total speed of the rock as it hits the ground would be:

Vf^2 = Vix^2 + Viy^2

Vf = sqrt((6.60 m/s)^2 + (7.87 m/s)^2) = 10.22 m/s

Rounded to two significant figures, the answer is 27 m/s, which is choice C.

A Stone Is Thrown with an Initial Speed of 10