Two food packets are thrown with the same velocity in the same direction with different angles of projection simultaneously. The angle of projection of one packet is `15^(@)`. At the same moment one boy starts running from rest from the point of projection with an acceleration of `10ms^(-2)` to catch them. If he caught one packet at a distance of 20 m and other packet in `1/2s` later the first packet, then the angle of projection of the second packet is
(Acceleration due to gravity, `g=10ms^(-2)`)

Options

A. `1/2sin^(-1)(25/32)`
B. `1/2sin^(-1)(8/9)`
C. `1/2 sin^(-1)(7/8)`
D. `1/2sin^(-1)(5/6)`

Answer

A

Explanation

According to the question, the given situation shown in the figure below

From second equation of the motion, `" "s_(A)=u_(A)t+1/2at^(2)["Given,"{:(S_(a)=20m), (a=10ms^(2)):}]`
`rArr " "20=1/2(10)t^(2) rArr t=2s`
Range of projectile `(R_(A))=(u_(A)^(2)sin2theta_(A))/g=20`
`" "[because {:("initial veocity,"), (" "u=0):}]`
`rArr " "u_(A)^(2)/10sin30^(@)=20`
`rArr " "u_(A)^(2)=200/(1//2) rArr u_(A)=20m//s`
Range of projectile for B,
`R_(B)=u_(B)t+1/2at^(2)=1/2 times 10 times (2+1/2 )^(2) " "(because u_(S)=0)`
`" "=5 times (5/2)^(2)=31.25m`
`because " "R_(B)=(u_(B)^(2)sin2theta_(B))/g`
`rArr 31.25=((20)^(2)sin2theta_(B))/10(becauseu_(A)=u_(B)=20m//s)`
`rArr sin2theta_(B)=312.5/(20)^(2)`
or `" "theta_(B)=1/2sin^(-1)(25/32)`