Question

A quarterback throws a football toward a receiver with an initial speed of 20 m/s at an angle of 30∘ above the horizontal. At that instant the receiver is 20 m from the quarterback. In (a) what direction and (b) with what constant speed should the receiver run in order to catch the football at the level at which it was thrown?

Answer 1

Answer:

a) In order to catch the ball at the level at which it is thrown in the direction of motion.

b)Speed of the receiver will be 7.52m/s

Explanation:

Calculating range,R= Vo^2Sin2theta/g

R= (20^2×Sin(2×30)/9.8 = 35.35m

Let receiver be(R-20) = 35.35-20= 15.35m

The horizontal component of the ball is:

Vox= Vocostheta= 20× cos30°

Vox= 17.32m/s

Time taken to coverR=35.35m with 17.32m/s will be:

t=R/Vox= 35.35/17.32

t= 2.04seconds

b)Speed required to cover 15.35m at 2.04seconds

Vxreciever= d/t = 15.35/2.04 = 7.52m/s

Answer 2

The speed of the receiver should run in order to catch the ball is 7.52 m/s.

Range can be calculated by the formula,

$\bold {R= \dfrac {Vo^2Sin2 \theta}{g}}$

Where,

Vo - Initial velocity

g - gravitational acceleration

Put the value in the formula,  

$\bold {R= \dfrac {(20^2\times Sin(2\times 30)}{9.8}} \\\\\bold {R = 35.35m}$    

The horizontal component of the ball is:  

$\bold {Vox= Vo \cos \theta} \\\\\bold {Vox= 20 \times cos30^o}\\\\\bold {Vox= 17.32m/s}$

Time taken to cover the range in 17.32m/s will be:  

$\bold {t= \dfrac {R}{Vox}}\\\\\bold { t = \dfrac {35.35}{17.32}}\\\\\bold {t= 2.04\ sec}}$

B) Speed required to cover 15.35m at 2.04seconds  

$\bold {Vx\ reciever= \dfrac d{t }}\\\\\bold {Vx\ reciever= \dfrac {15.35}{2.04} }\\\\\bold {Vx\ reciever= 7.52m/s}$

Therefore, the speed of the receiver should run in order to catch the ball is 7.52 m/s.

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