Question

A body is projected vertically upwards. If t1 and t2 be the times at wch it is at height h above the point of projection wle ascending and descending respectively. then h is,

Answer 1

u = velocity of projection (upward).
a = acceleration (downward) = - g
s = displacement after time t = h
Applying the relation s = u t + (1/2) a t², we have
h = u t - (1/2) g t²
=> 2h = 2ut - gt²
=> gt² - 2ut + 2h = 0 .
This is a quadratic equation in t which shows that for a given h there are two values of t.
Multiplying the above equation by g through out we get
g²t² - 2ugt + 2gh = 0
=> (gt)² - 2(gt)(u) + u² - u² + 2gh = 0
=> (gt - u)² = u²- 2gh
=> gt - u = ± √(u²- 2gh)
=> gt = u ± √(u²- 2gh)
=> t = {u ± √(u²- 2gh)} / g
=> t1 = {u + √(u²- 2gh)} / g and t2 = {u - √(u²- 2gh)} / g
Multiplying, (t1)(t2) = {u² - (u² - 2gh)} / g² = 2gh / g²
=>(t1)(t2) = 2h / g
=> h = (1/2)g(t1)(t2) 

Answer 2

Answer:

$\frac{g}{2}$$(\frac{t_{1+t_{2} } }{2} )^{2}$

Explanation:

Since the body is projected vertically upward do initial velocity i.e u =0 m/s.

time = $t_{2} - t_{1}$

Therefore half time is ($\frac{t_{2} - t_{1} }{2}$)

We know,

s = ut + (1/2)a$t^{2}$

by putting the values

=> s = 0 +(1/2) g$t^{2}$

putting the value of t

=> s = (1/2)g {$t_{1}$ -($t_{2} - t_{1}$)/2}^2

=> s = $\frac{g}{2}$$(\frac{t_{1}+t_{2} }{2}) ^{2}$ Ans

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