Question

From the top of the tower of height 50m a ball is projected upwards with a speed of 30m/s at an angle of 30. to the horizontal then calculate

a) maximum height frm ground.

b) at wt distance frm the foot of tower does the projectile hit the ground.

c) time of flight.

Answer 1

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.

 

Answer 2

Answer:

Explanation:

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