Question

A ball is thrown with a velocity of 7root 2 m/s at an angle of 45degree with the horizontal. Itjust clears two vertical poles of height 90 cm each.find the separation between the poles.

Answer 1

Answer on Question #60387-Physics-Mechanics-RelativityA ball is thrown with a velocity of 7 root 2 m/s at an angle of 45 degree with the horizontal. It just clears two vertical poles of height 90 cm. Find the separation between the poles. Take g =9.8SolutionLet t be the time after which the ball is at the top of the poles.푦(푡)=0.9푚푣푦(0)=푣푠푖푛45º=7√21√2=7푚푠.Now, 푦(푡)=푣푦(0)푡+푎푦푡220.9=7푡–(9.8)푡24.9푡2–7푡+0.9=0On solving, we get푡1=17푠푎푛푑푡2=97푠.Hence the ball is at A after 17s and at B after 97s.푂푃=푣푥푡1=717=1푚푂푄=푣푥푡2=797=9푚푃푄=9−1=8

Answer 2

Answer:

The separation between the poles is 2 m.

Explanation:

Given that,

Velocity$v =7\sqrt{2}\ m/s$

Height h = 90 cm = 0.9 m

Angle = 45°

let t be the time after which the ball is at the top of the poles.

The vertical velocity is

$v_{y}=v_{y}\sin\theta$

$v_{y}=7\sqrt{2}\times\sin45^{\circ}$

$v_{y}=7\sqrt{2}\times\dfrac{1}{\sqrt{2}}$

$v_{y}=7\ m/s$

Now, we need to calculate the time

Using equation of projectile

$s_{y}=u_{y}t+\dfrac{1}{2}gt^2$

$0.9=7t-\dfrac{1}{2}\times9.8t^2$

$4.9t^2-7t+0.9=0$

After solving

The value of t is $\dfrac{1}{7}\ sec$ and $\dfrac{9}{7}\ sec$ .

A ball at First pole after $\dfrac{1}{7}\ sec$ and at second pole after $\dfrac{9}{7}\ sec$ .

Now, the separation between the poles

The distance of the first pole from the origin

$D = v_{x}\times t$

$D=7\sqrt{2}\times\dfrac{1}{\sqrt{2}}\times\dfrac{1}{7}$

$D=1\ m$

The distance of the second pole from the origin

$D'=7\sqrt{2}\times\dfrac{1}{\sqrt{2}}\times\dfrac{9}{7}$

$D'=9\ m$

The separation between the poles is

$d=D'-D$

$d=9-7$

$d=2\ m$

Hence, The separation between the poles is 2 m.