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Q42E

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College Physics (Urone)
Found in: Page 125

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Short Answer

Suppose a soccer player kicks the ball from a distance 30 m toward the goal. Find the initial speed of the ball if it just passes over the goal, 2.4 m above the ground, given the initial direction to be 40º above the horizontal.

The initial velocity of the ball will be 18.1 m/s.

See the step by step solution

Step by Step Solution

Step 1: Given data

  • The change in the displacement in the y-direction is 2.4 meters.
  • The acceleration in the above case is -9.8 m/s2.

Step 2: Initial velocity of the ball

Figure: The motion of the ball

The initial velocity in the x-direction can be caculated as:

cosθ=ahcos40=VixViVix=Vicos40°

The initial velocity in the x-direction is the above equation.

The initial velocity in the y-direction can be caculated as:

sinθ=ohsin40=ViyViViy=Visin40

The initial velocity in the y-direction is above the equation.

The velocity in the x frame will be constant at all times. Considering the initial velocity, we can write:

V=xtVix=30tt=30Vicos40

The time taken is used from the above equation.

From the equation of motion:

Y=Viyt+12ayt2

By putting the values in the above equation, we get:

2.4=Visin4030Vicos40+12-9.830Vicos4022.4=25.2-4410Vi2×0.5874410Vi2×0.587=25.2-2.4Vi2=751022.8Vi2=329Vi=18.1 ms

Therefore, the initial velocity will be 18.1 m/s.

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