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.
