Question

A particle is thrown from top of a building of height 50m with a velocity 30m/s at an angle of 30degree above the horizontal then the distance from the base of building to the point where the particle hits the ground is...

Answer 1

Explanation:

Along the y-direction,

Initial velocity = u sinθ

Acceleration due to gravity = -9.8 m/s⁻²

∴ S = u_yt + \frac{1}{2} at^2

⇒ 50 = -usinθt + 1/2 × 10 × t²

⇒ -15t + 5t² = 50

⇒ t² - 3t - 10 = 0

∴ t² - 5t + 2t - 10 = 0

⇒ t(t - 5) + 2(t - 5) = 0

∴ (t + 2)(t - 5) = 0

⇒ t = -2 and t = 5

Hence, Time is 5 seconds.

Therefore, distance along horizontal = ucosθ × t = 30 × √3/2 × 5 = 75√3 m.

For Maximum height,

H = u²sin²θ/2g

= 900/80

= 45/4 m.

= 11.25 m.

Total maximum height = 50 + 11.25 = 61.25 m.