a person standing on the top of a cliff 30 m high has to throw a packet to his friend standing on the ground 40 m horizontal away. if he throws the packet directly aiming at the friend with aspeed of 125/3 m/s. packet fall at a distance 20/@ m from the friend. here @is an integer find @ (use g=10 m/s)
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Final Answer :@= 3
That is,
Packet falls 20/3 m. before the friend.
Steps and Understanding :
1) We will find time taken by packet to reach the ground by vertical equation of motion.
Let the angle made by speed with vertical be theta.
S(y) = 30m
a(y) = 10m/s^2
u(y) =125/3 cos theta
See pic 1 for calculation.
2) We will find horizontal distance from cliff by horizontal equation of motion.
a(x) = 0
u(x) = 125/3 sin (theta)
S(x) = x (say) horizontal distance from. cliff.
3) Find distance from friend d = (40- x) as shown in pic.
That is,
Packet falls 20/3 m. before the friend.
Steps and Understanding :
1) We will find time taken by packet to reach the ground by vertical equation of motion.
Let the angle made by speed with vertical be theta.
S(y) = 30m
a(y) = 10m/s^2
u(y) =125/3 cos theta
See pic 1 for calculation.
2) We will find horizontal distance from cliff by horizontal equation of motion.
a(x) = 0
u(x) = 125/3 sin (theta)
S(x) = x (say) horizontal distance from. cliff.
3) Find distance from friend d = (40- x) as shown in pic.
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Answer:
20/3
Explanation:
3 aavse
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