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
The speed of sound at

is approximately v=343 m/s. The distance covered by the sound wave is

And the time it takes is

Now we want to find how far the light travels during this time. Light travels at speed

, therefore the distance it covers during this time is
1 watt = 1 joule/sec
2,000 watts = 2,000 joules/sec
(2,000 joule/sec) x (120 sec)
= (2,000 x 120) (joule-sec/sec)
= 240,000 joules .