Physics, asked by apd04, 1 year ago

A ball thrown vertically up from the top of a tower reaches the ground in 12 s. Another ball thrown vertically downwards from the same position with the same speed takes 4 seconds to reach the ground. Find the height of the tower.

Take G =10 m/s^2​

Answers

Answered by amitnrw
30

Distance = ut + (1/2)at^2

Height of tower = distance for downward thrown ball


H = 4u + (1/2)10(4)^2

H = 4u + 5×16

H = 4u + 80


Out of 12 sec for upwards thrown ball 4 secs to go below tower and rest 8 secs half to go upward and half to come to tower peak again

So aftet 4 secs ball will be at its peak

Ball thrown up ward speed at top = 0

V = u + at

a = -g as going upward

0 = u + (-10)×4

u = 40 m/s


H = 4×40 + 80

H = 240m



apd04: thank you very much sir
apd04: but sir in this problem you can't tell that when the second ball is moving from the height of tower and then coming back in the height of the tower , it will be in the same speed of the first ball so that it will reach the ground in 4 seconds. it can be slower or faster than the ball and the time will be different
amitnrw: Ball thrown upward is with same speed as ball thrown upward let say speed = u m/s . You know very well speed at top = 0 0 = u - gt so t = u/g (time taken to reach at top from tower). as magnitude of acceleration is same going upward or downward . so time taken from top to tower will be same as time taken from tower to top (u/g).
amitnrw: Now speed at top = 0 not after (u/g) time bal will be again at tower height so speed at tower height = u + gt = 0 + g(u/g) = u (same as the speed of ball thrown upward and downward) . Hope it clears your doubt
Answered by lovedeepgehal
2

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