cherishdisah
Answered

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A ball is thrown up a smooth inclined plane with initial velocity of 54km/h. If the inclination of the plane is 30°, find
I) the time taken for the object to reach its maximum height.
II) the maximum height reached.
[Take g=10m/s²]​

Sagot :

chibabykalu16

Answer:

I) 3 seconds

ii) 22.5 meters

Explanation:

Given:

  • Gravity (g) = 10m/s²
  • Initial velocity (u) = 54 km/h = 15 m/s (the required scale)
  • Projected at angle 30° = θ

I) Time taken to reach maximum height:

The ball will reach its maximum height when its vertical component of velocity is zero. We can use the following kinematic equation:

v = u + at

where:

v = 0 (final velocity at maximum height)

u = 54 km/h = 15 m/s (initial velocity)

a = -g sin(30°) = -10 m/s² × 0.5 = -5 m/s² (acceleration down the inclined plane)

t = ? (time to reach maximum height)

Rearranging the equation to solve for t:

t = -u / a

= -15 m/s / (-5 m/s²)

= 3 s

So, it takes 3 seconds for the ball to reach its maximum height.

II) Maximum height reached:

We can use the following kinematic equation:

s = ut + (1/2)at²

where:

s = ? (maximum height)

u = 15 m/s

t = 3 s

a = -5 m/s²

Rearranging the equation to solve for s:

s = ut + (1/2)at²

= (15 m/s × 3 s) + (1/2 × -5 m/s² × 3²)

= 45 m - 22.5 m

= 22.5 m

So, the maximum height reached is 22.5 meters.

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