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A football is thrown at an angle of 30.° above the horizontal. The magnitude of the horizontal

component of the ball’s initial velocity is 13.0 meters per second. The magnitude of the vertical

component of the ball’s initial velocity is 7.5 meters per second. [Neglect friction.]


The football is caught at the same height from which it is thrown. Calculate the total time the football

was in the air. [Show all work, including the equation and substitution with units.]

Respuesta :

Answer:

1.53 s

Explanation:

Initially vertical component of velocity of the ball, uy = 7.5 m/s

Net displacement is vertical direction is zero, Δy =0

Use second equation of motion:

Δy = uy t + 0.5 a t²

Here, acceleration a = -g                           (g =9.8 m/s²)

Substitute all the values and solve for g

0 = 7.5 t -0.5 (9.8)t²

7.5 t = 4.9 t²

t = 1.53 s

The total time when football was on the air is 1.53 s.

From second equation of motion:  

Δy = Uy t + 0.5 a t²

Where,

Δy -  Net displacement is vertical direction = 0

Uy - initial velocity of the ball = 7.5 m/s

a - acceleration = -g   = 9.8 m/s²  

t - time taken = ?  

Put the values in the formula,  

0 = 7.5 t -0.5 (9.8) t²  

7.5 t = 4.9 t²  

t = 1.53 s

Therefore, the total time when football was on the air is 1.53 s.

To know more about  second equation of motion:

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