Mechanics> A body is projected vertically upwards. I...

A body is projected vertically upwards. If t1 and t2 be the times at which it is at height h above the point of projection while ascending and descending respectively, then h is, (1)1/2gt1t2 (2)gt1t2 (3)2g(t1+t2) (4)4g(t1+t2)

Girish Pimpalgaonkar , 8 Years ago

Grade 12

6 Answers

Croma Tab

t₁is the time the particle takes to reach h.

so,

h = ut₁- ½ gt₁² $\Rightarrow$ u = (h + ½ gt₁²) / t₁ (equation 1)

t₂ is the time taken by the particle to reach the same point again.

so,

h = ut₂- ½ gt₂²

substituting 1 in the above equation:

h = (h + ½ gt₁²) t₂/t₁ – ½ gt₂²

solving

h = ½ gt₁t₂

Last Activity: 8 Years ago

nithishagattu

Let t₁be the time where particle reach height " h" and let t₂ be the time where particle attains same height "h" again.

Time of flight T=t₁ +t₂

h=(usinθt)+1/2gt² ; u is velocity of projection.

gt²-2usinθt+2h=0

the roots of this equation are t₁and t₂

t₁t₂=2h/g, t₁+t₂=2usinθ/g [for ax²+bx+c=0,sum of the roots=-b/a and product of the roots=c/a]

By solving these equations we get h=1/2gt₁t₂

Last Activity: 8 Years ago

Aakash chalise

a body is projected vertically upwards from the ground.It is found at the same elevation at t=3 and t=7s after projection.Find the projection speed.

Last Activity: 7 Years ago

Aakash chalise

a body is projected vertically upwards from the ground.It is found at the same elevation at t=3 and t=7s after projection.Find the projection speed.

Answer.

U=g÷2(t1+t2)

=5×(3+7)

=5×10

=50m/s.

U=g÷2(t1+t2)

=5×(3+7)

=5×10

=50

Last Activity: 7 Years ago

Pranav Kumar Varshney

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.

Solving the equation for t using Quadratic formula(x=(-b± √b²-4ac)/2a)

t=(2u± √4u²-8gh)/2g

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)

Last Activity: 6 Years ago

Kushagra Madhukar

Dear student,

Please find the answer to your problem below.

we have,

u = velocity of projection (upward).

a = acceleration (downward) = – g

s = displacement after time t = h

Applying the relation s = ut + (1/2)at², we have

h = ut – (1/2)gt²

or, gt² – 2ut + 2h = 0

Now, since it is a quadratic equation, and its roots are t₁ and t₂

Hence, Product of roots = c/a = 2h/g

or, t₁t₂ = 2h/g

Hence, h = (1/2)gt₁t₂

Hope it helps.

Thanks and regards,

Kushagra

Last Activity: 5 Years ago

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Mechanics> A body is projected vertically upwards. I...

A body is projected vertically upwards. If t1 and t2 be the times at which it is at height h above the point of projection while ascending and descending respectively, then h is, (1)1/2gt1t2 (2)gt1t2 (3)2g(t1+t2) (4)4g(t1+t2)

Girish Pimpalgaonkar , 8 Years ago

Croma Tab

nithishagattu

Aakash chalise

Aakash chalise

Pranav Kumar Varshney

Kushagra Madhukar

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