Question

initial speed of an object thrown horizontally=25m/s
it strikes the ground after covering a horizontal distance of 75m (g=10m/s)

a)how high is the building?
b)what is the speed of the object when it hit the ground?

THE ONE WHO EXPLAINS CLEARLY WILL BE MARKED BRAINLIEST!

Answer 1

As it is given that, the object it initially thrown horizontally, the initial velocity of 25 m/s will be provided by the horizontal component of velocity alone, and the vertical component of velocity will be 0.

So,

Given:-

• uₓ = 25 m/s
• uᵧ = 0 m/s
• sₓ = 75 m

To find:-

• Height of building = H
• Velocity while hitting the ground = vₙₑₜ

Answer:-

In horizontal direction:-

sₓ = uₓt + (1/2)aₓt²

Since there is no acceleration in horizontal direction, aₓ = 0 m/s². Putting available values,

→ 75 = (25 × t) + (1/2 × 0 × t²)

→ 75 = (25 × t) + 0

→ t = 75/25

→ t = 3 seconds

In vertical direction:-

sᵧ = uᵧt + (1/2)aᵧt²

Acceleration (aᵧ) will be (-g) since it is acting in downward direction, and displacement (sᵧ) = -H since it is also in downward direction. Putting available values,

→ -H = (0 × 3) + (1/2 × -g × 3 × 3)

→ -H = 1/2 × (-10) × 3 × 3

→ H = 45 metres Ans.

Again,

In horizontal direction:-

vₓ = uₓ + aₓt

→ vₓ = 25 + (0 × 3)

→ vₓ = 25 m/s ----( 1 )

In vertical direction:-

vᵧ = uᵧ + aᵧt

→ vᵧ = 0 + (-g × 3)

→ vᵧ = - 30 m/s ----( 2 )

Net velocity while hitting the ground:-

vₙₑₜ = √[vₓ² + vᵧ²]

→ vₙₑₜ = √[ (25)² + (-30)² ] [From ( 1 ) and ( 2 )]

→ vₙₑₜ = √[ 625 + 900 ]

→ vₙₑₜ = √[1525]

→ vₙₑₜ = 39.051 m/s Ans.