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ID: 1578193 • Letter: T
Question
t.cns.utexas.edu/student/images/problem_pdf? courseuserassignment-35800387 012 10.0 points A ball is thrown from the top of a building upward at an angle of 50° to the horizontal and with an initial speed of 24 m/s. The ball is thrown at a height of 54 m above the ground and hits the ground 87.7659 m from the base of the building. : .,_ " hand height " ground level What is the speed of the ball just before it strikes the ground? The acceleration due to gravity is 9.8 m/s2 . Answer in units of m/s.Explanation / Answer
Given,
Initial velocity Vo=24 m/s
Angle =50o,
height=54 m above grd,
Voy=Vo*sin()18.38 m/s
Vox=Vo*cos()15.42 m/s
a) -54=Vo*sin(50o)-1/2*g*t2
Total time of flight t=5.695 s
b) final vertical component of velocity
Vy=voy-g*t=18.38-9.8*5.695 -37.43 m/s
Total speed =(15.422+37.432)40.48 m/s
speed of ball just before hitting grd is 40.48 m/s
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