Question

An object is projected with speed 5 m/s from top of a tower of height
H = 50 m. Angle of projection is 30° upward from horizontal. Calculate the
(1) Time t at which the velocity vector is perpendicular to initial velocity
vector.
(ii) Time taken by it to hit the ground and angle from the horizontal with
which it hits the ground.
(iii) Magnitude of displacement when the object hits the ground.
plz
answer me ASAP, it's an emergency
I will mark u brainleist plz​

Answer 1

Answer:

Along the y-direction,

Initial velocity =  u sinθ

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

hope this will help u:)

Answer 2

Answer:

See the attachments

Explanation: