Question #296450

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.

1
Expert's answer
2022-02-11T08:47:26-0500

a. We have to solve the equation h = h0 + V0yt - ½gt² when h = 0 m as the ball hits the ground, h0 is the height of the building, and the velocity will be V0y = 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+10t4.9035t2=0.45+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:

V0x=V0cos(30°)=(20ms)(32)=17.32msV0y=V0sin(30°)=(20ms)(12)=10msV_{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:

Vy=V0ygtVy=[10(9.807)(4.216)]ms=31.346msV_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=Vy2+Vx2V=(31.346ms)2+(17.32ms)2V=35.81msV=\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=V0xt=(17.32ms)(4.216s)=73.02md = 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.


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