Question

A ball projected with a velocity of 10m/s at angle of 30 with horizontal just clears two vertical poles each of height Im. Find separation between the poles.​

Answer 1

Answer:

here separation could be considered as range

r = 10×10×√3/2×10

= 5√3

Answer 2

Answer:

Since the projection is horizontal

the Y component of velocity = u Sinθ

the initial velocity of ball is 10 m and angle of projection is 30° (  given )

so , v = u sinθ

= 10 sin 30°

= 10 ×$\frac{1}{2}$ = 5m

using the second equation of motion with respect to gravity

h = ut + $\frac{1}{2}$ g$t^{2}$

the height of poles is given 1 m

substituting the values in equation of gravity we get

= 10 sin 30 t - $\frac{1}{2}$(- 10) $t^{2}$

 5$t^{2}$ - 5t +1 =0

solving by using quadratic equation we get two values of t

t = 0.72 sec. and t = 2.76 sec.

speed =$\frac{distance}{time}$

vCosθ = $\frac{d}{t_{2} -t_{1} }$

separation = vCos θ ( 2.76 - 0.72 )

 10 cos 30 × 2.04

10×$\frac{\sqrt{3} }{2}$×2.04

= 5 ×1.73×2.04

= 17.7 m

separation of poles = 17.7m

For projectile motion visit :

https://brainly.in/question/15793946

formula to solve projectile visit:

https://brainly.in/question/12720733

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