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Sia ? 6 years, 6 months ago
Let body takes T sec to reach maximum height.
We know, v = u - gT
At the highest point, v = 0
T = {tex}\frac ug{/tex} ………… (i)
Velocity attained by Body in (T - t) s,
v = u - g(T - t)
= u - gT + gt = u - g{tex}\frac ug{/tex} + gt
or v = gt......(ii)
Now, Distance travelled in last t sec of its ascent
s = (gt)t -{tex}\frac 12{/tex}gt2 = {tex}\frac 12{/tex}gt2
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Coefficient of elasticity is the ratio of the applied stress to the change in shape of an elastic body.
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Refer to the similar ques:
A bomb is dropped from an aeroplane when it is directly above target from height 1000 m, aeroplane is moving with velocity 500 kmph find how much distance will bomb miss target ?
Time taken to reach the ground = √(2H/g)
= √(2 × 1000 / 10)
= √(200)
= 10√2 seconds
Distance = Velocity × Time
= (500 × 5/18 m/s) × 10√2 s
= 1964.1 m
= 1.964 km
The bomb wil miss the target by distance of 1.964 km

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Tripti Rawat 6 years, 6 months ago
Magnitude of vector A = {tex}\sqrt{(2)^2 + (6)^2 + (1)^2}{/tex}={tex}\sqrt{4+36+1}{/tex}={tex}\sqrt{41}{/tex}
Magnitude of vector B = {tex}\sqrt{(4)^2 + (2)^2 + (-11)^2}{/tex}={tex}\sqrt{16+4+121}{/tex}={tex}\sqrt{141}{/tex}
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