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Lunna [17]
3 years ago
15

A particle was moving in a straight line at 172.8 km/hr. If it decelerated over 120 meters to come to rest, find the time taken

to cover this distance.​
Physics
1 answer:
Bezzdna [24]3 years ago
8 0

Answer:

v=s/t

s=vt

t=s/v

t=(120×10‐³)/172.8

(the distance meters has been changed to kilometres)

t=1/1440 hrs

Given ,

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Answer:

H = 45 m

Explanation:

First we find the launch velocity of the ball by using the following formula:

v₀ = √(v₀ₓ² + v₀y²)

where,

v₀ = launching velocity = ?

v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s

v₀y = Vertical Component of Launch Velocity = 30 m/s

Therefore,

v₀ = √[(15 m/s)² + (30 m/s)²]

v₀ = 33.54 m/s

Now, we find the launch angle of the ball by using the following formula:

θ = tan⁻¹ (v₀y/v₀ₓ)

θ = tan⁻¹ (30/15)

θ = tan⁻¹ (2)

θ = 63.43°

Now, the maximum height attained by the ball is given by the formula:

H = (v₀² Sin² θ)/2g

H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)

<u>H = 45 m</u>

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