Name: Motion With Variable Acceleration
Uploaded: 2019-07-19
Duration: 10 min 54 s
Description: You may be aware of the SUVAT equation, but do you think this can be applicable for all types of motion? No, what are those motion? Watch this to find out.

Question

A player kicks a football at an angle of

45 °

with an initial speed of

20 m s^{- 1}

. A second player on the goal line

60 m

away in the direction of kick-starts running to receive the ball at that instant. Find the constant speed of the second player with which he should run to catch the ball before it hits the ground.

(g = 10 m s^{- 2})

Accepted Answer

The football is kicked with a speed of $v = 20 m s^{- 1}$ at an angle of $45 °$ . From this we can find the horizontal velocity and the vertical velocity of the ball at the moment it was kicked.

Vertical velocity of the ball is $v_{y} = v \sin (45 °) = 20 \frac{1}{\sqrt{2}} = 10 \sqrt{2} m s^{- 1}$ .

Horizontal velocity of the ball is $v_{x} = v \cos (45 °) = 20 \frac{1}{\sqrt{2}} = 10 \sqrt{2} m s^{- 1}$ .

Time of flight for the ball moving in projectile motion is given by,

$T = \frac{2 v_{y}}{g} = \frac{2 \times 10 \sqrt{2}}{10} = 2 \sqrt{2} s$ .

Now the horizontal range of the ball is

$R = \frac{2 v_{x} v_{y}}{g} = \frac{2 \times 10 \sqrt{2} \times 10 \sqrt{2}}{10} = 40 m$ .

The second player is $60 m$ away from the point where the ball was kicked. The distance this player needs to cover is the remaining distance from its position to where the ball will drop. This distance is $d = 60 - 40 = 20 m$ .

Also, he needs to reach there before the ball drops so the time taken by him is equal to the time of flight. The time taken by the second player is $t = 2 \sqrt{2} s$ .

Speed of the second player is $\frac{d}{t} = \frac{20}{2 \sqrt{2}} = 5 \sqrt{2} m s^{- 1}$ .

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