A man stands on the roof of a building of height 13.7m and throws a rock with a velocity of magnitude 28.0m/s at an angle of 27.5 degrees above the horizontal.
A. calculate the maximum height above the roof reached by the rock. Answer in m
B. Calculate the magnitude of the velocity of the rock just before it strikes ground. Answer in m/s
C. Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground. Answer in m
A man stands on the roof of a building of height 13.7 m and throws a rock with a velocity of magnitude 28.0 m/s at an angle of 27.5 degrees above the horizontal.
The maximum height that the projectile reaches is given by (v*sin x)^2/2*g where v is the initial velocity and x is the angle made with the horizontal. Substituting the values from the problem the maximum height is (28*sin 27.5)^2/19.6 = 8.528 m. But the rock is thrown from the top of a building at a height of 13.7 m. This gives the maximum height as 22.22 m
The magnitude of the velocity when it completes the projectile path and reaches the height it was thrown from is 28.0 m/s at an angle of 27.5 degrees with the horizontal. Here, the horizontal component of the velocity is 28*cos 27.5 = 24.5 m/s The vertical component is 28*sin 27.5, but as the rock falls another 13.7 m it increases to v, where v^2 = (28*sin 27.5)^2 + 2*9.8*13.7 = 20.87 m/s . The magnitude of the velocity is sqrt(20.887^2 + 24.5^2) = 32.19 m/s
The range of the rock till it reaches the height it was thrown from is v^2*(sin 2x)/g = 65.53 m. The time taken by the rock to fall 13.7 m is t where 13.7 = 28*sin 27.5*t + (1/2)*9.8*t^2
Solving, the equation t = 0.6070 s. IN this time the horizontal distance covered is 24.5*0.6070 = 14.88 m
This gives the distance from the base of the building where the rock hits the ground as 80.41 m