Question #107143

Show complete solution: Use the correct number of significant figures.


1. A soccer ball is kicked from the ground with an initial speed of 19.6 m/s at an upward angle of 46.4˚. A player 50.6 m away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground? Neglect air resistance.


Answer: _________

( the tolerance is +/-2%)


2. A batter hits a pitched ball when the center of the ball is 1.27 m above the ground.The ball leaves the bat at an angle of 45° with the ground. With that launch, the ball should have a horizontal range (returning to the launch level) of 100 m.


(a) Does the ball clear a 7.41-m-high fence that is 90.0 m horizontally from the launch point?


(b) At the fence, what is the distance between the fence top and the ball center?

Expert's answer

As per the given question,

Kicked speed of the ball (u)=19.6 m/sec

Upward angle (θ)=46.4∘(\theta)=46.4^\circ

distance between the player and the ball (d)=50.6m

gravitational acceleration (g)=9.8m/sec2(g)=9.8 m/sec^2

Now,

Time of flight of the ball=2usin⁡θg\dfrac{2u\sin \theta}{g}

=2×19.6×sin⁡46.49.8=\dfrac{2\times19.6\times \sin46.4}{9.8}

=2.896sec∼2.9sec=2.896sec\sim2.9 sec

horizontal range of the projectile d=u2sin⁡2θgd=\dfrac{u^2\sin2\theta}{g}

=19.62sin⁡(2×46.4)9.8=\dfrac{19.6^2\sin(2\times 46.4)}{9.8}

=39.15m∼39.2m=39.15m \sim 39.2m

total distance covered by player = 50.6m−39.2=11.4m50.6m-39.2=11.4m

Ball and player both will be at the same position at same time.

Let the speed of the player be v,

So, v=11.42.9m/secv=\dfrac{11.4}{2.9}m/sec

v=3.93m/secv=3.93 m/sec


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