Question #15613

A bal thrown straight upward returns to its original level in 2.20 s. A second ball is thrown at an angle of 42.0 degrees above the horizontal. What is the initial speed of the second ball if it also returns to its original level in 2.20 s?

Expert's answer

The vertical component V of the ball's initial velocity I is

V = S*sin(42°).& (1)

The vertical component of the ball's velocity turns to zero at the highest point of the flight. Also, the highest point of the flight is reached at the half of the flight time. So,

V = gT/2 = 9.8[m/s²]*2.20[s]/2 = 10.78 m/s.

As follows from (1),

I = V/sin(42°) = 10.78[m/s]/sin(42°) ≈ 16.11 m/s.

So, the initial velocity of the ball was about 16.11 m/s.

V = S*sin(42°).& (1)

The vertical component of the ball's velocity turns to zero at the highest point of the flight. Also, the highest point of the flight is reached at the half of the flight time. So,

V = gT/2 = 9.8[m/s²]*2.20[s]/2 = 10.78 m/s.

As follows from (1),

I = V/sin(42°) = 10.78[m/s]/sin(42°) ≈ 16.11 m/s.

So, the initial velocity of the ball was about 16.11 m/s.

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