Question

1) A stone is thrown with an initial speed u (m/s) . The height of the stone's trajectory 1 point
above the ground is, H(t)= -5(t)^2+1/2ut
(here t is the time of flight). If the highest point in air that the stone can
reach is 5m above the ground, then calculate the initial speed u.​

Answer 1

Given:

A stone is thrown with an initial speed u (m/s) . The height of the stone's trajectory above the ground is, H(t)= -5(t)^2+1/2ut (here t is the time of flight). The highest point in air that the stone can reach is 5m above the ground.

To find:

Initial velocity ?

Calculation:

$\therefore \sf H = - 5 {t}^{2} + \dfrac{1}{2} ut$

For maxima , 1st order derivative should be zero and 2nd order derivative must be less than zero.

$\therefore \sf \dfrac{dH}{dt} = \dfrac{d \bigg(- 5 {t}^{2} + \dfrac{1}{2} ut \bigg)}{dt}$

$\implies \sf \dfrac{dH}{dt} = - 10t + \dfrac{u}{2} = 0$

$\implies \sf - 10t + \dfrac{u}{2} = 0$

$\implies \sf t = \dfrac{u}{20}$

Now, 2nd order derivative:

$\implies \sf \dfrac{{d}^{2} H}{d{t}^{2} } = \dfrac{d \bigg( - 10t + \dfrac{u}{2} \bigg)}{dt}$

$\implies \sf \dfrac{{d}^{2} H}{d{t}^{2} } = - 10$

$\implies \sf \dfrac{{d}^{2} H}{d{t}^{2} } < 0$

So, maxima is obtained at t = u/20.

Putting value of t in eq.(1):

$\therefore \sf H = - 5 {t}^{2} + \dfrac{1}{2} ut$

$\implies\sf H = - 5 { (\dfrac{u}{20} )}^{2} + \dfrac{1}{2} u( \dfrac{u}{20} )$

$\implies\sf H = - \dfrac{ {u}^{2} }{80} + \dfrac{ {u}^{2} }{40}$

$\implies\sf H = \dfrac{ {u}^{2} }{80}$

Now, this max height is 5m:

$\implies\sf 5= \dfrac{ {u}^{2} }{80}$

$\implies\sf {u}^{2} = 5 \times 80$

$\implies\sf {u}^{2} = 400$

$\implies\sf u = 20 \: m {s}^{ - 1}$

So, Initial velocity is 20 m/s.

Answer 2

Given : A stone is thrown with an initial speed u (m/s) . The height of the stone's trajectory above the ground is, H(t)= -5(t)²+(1/2)ut

highest point in air that the stone can  reach is 5m above the ground,

To Find :

Step-by-step explanation:

H(t) = -5t²  + (1/2)ut

dH/dt  = - 10t  + (1/2)u

dH/dt  = 0

=>  - 10t  + (1/2)u = 0

=> u = 20t

or t = u/20

d²H/dt²  =  -10 < 0

Hence maximum height at  t =  u/20

H(t) = -5t²  + (1/2)ut

= -5t² + (1/2)20t*t

= - 5t²  + 10t²

= 5t²

highest point in air that the stone can  reach is 5m above the ground,

=> 5t² = 5

=> t² = 1

=> t = 1  

u = 20t  = 20  

initial speed = 20 m/s

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