Answer to Question #296450 in Mechanics | Relativity for yan

Question

A stone is thrown upward from the top of a building at an angle of 30.0° to the

horizontal and with an initial speed of 20.0 m/s. The point of release is 45.0 m above the

ground.

a. How long does it take to hit the ground?

b. Find the stone’s speed impact.

c. Find the horizontal range of the stone.

Answer 1

a. We have to solve the equation h = h₀ + V_0yt - ½gt² when h = 0 m as the ball hits the ground, h₀ is the height of the building, and the velocity will be V_0y = V₀•sin(30°)=(20 m/s)(½)=10 m/s. Using that information, the quadratic equation that has to be solved for the time t is $45+10t-4.9035t^2= 0.$ Solving that we find the time as t = 4.216 s.

b. The velocity will be the combination of the velocity for the x and y axis. First:

$V_{0x}= V_{0}\cos(30°)=(20\frac{m}{s})(\frac{\sqrt{3}}{2})=17.32\frac{m}{s} \\ V_{0y}= V_{0}\sin(30°)=(20\frac{m}{s})(\frac{1}{2})=10\frac{m}{s}$

Then we have to consider that the new velocity for the y axis has to be calculated while the other velocity remains constant:

$V_y=V_{0y}-gt \\ V_y=[10-(9.807)(4.216)]\frac{m}{s}=-31.346\frac{m}{s}$

The total velocity will be $V=\sqrt{V_y^2+V_x^2} \\ V=\sqrt{(-31.346\frac{m}{s})^2+(17.32\frac{m}{s})^2} \\ V= 35.81\frac{m}{s}$

c. The distance that the stone travels through the x axis will be equal to $d = V_{0x}t= (17.32\frac{m}{s})(4.216\,s)=73.02\,m$

In conclusion, the stone travels a time of 4.216 s, hits the ground with a velocity of 35.81 m/s, and hits the ground at a distance of 73.02 m from the place where the stone was thrown.