Question: A ballerina leaps at an angle of 75 degrees (Above horizontal) with an

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Question: A ballerina leaps at an angle of 75 degrees (Above horizontal) with an initial speed of 4.5ms -1 A) Calculate the vertical component of the ballerinas initial velocity B) Calculate the time the ballerina spends in the air before landing C) Calculate her horizontal displacement when she lands D) A Second ballerina leaps directly upward (launch angle 90 degrees above horizontal) at exactly the same time as the first. What must the second ballerinas speed be if she lands at the same time as the first ballerina? Please explain to help me study Question: A ballerina leaps at an angle of 75 degrees (Above horizontal) with an initial speed of 4.5ms -1 A) Calculate the vertical component of the ballerinas initial velocity B) Calculate the time the ballerina spends in the air before landing C) Calculate her horizontal displacement when she lands D) A Second ballerina leaps directly upward (launch angle 90 degrees above horizontal) at exactly the same time as the first. What must the second ballerinas speed be if she lands at the same time as the first ballerina? Please explain to help me study

Explanation / Answer

The initial speed of ballerina u = 4.5 m/s
The angle made with the horizontal theta = 75 degree a) The vertical component of the initial speed vy = u sin theta vy = 4.5 m/s *sin 75 vy = 4.35 m/s b) The time spent before landing is T = 2u sin theta /g
T = 2*4.5 m/s*sin75/9.8 m/s^2 T = 0.887 s c) the horizontal displacement, R = u^2*sin2(theta)/g R = 4.5m/s^2*sin150/9.8 m/s^2 R = 1.033 m d)As both spent the same time in air, the second ballerina has to spend a time of 0.887 s in the air before it reaches the ground The time of flight T = 2u/g 0.887 s = 2*u/9.8m/s^2 u = 4.32 m/s

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