The maximum height reached by the ball is 99.2 m
Explanation:
When the ball is thrown straight up, it follows a free fall motion, which is a uniformly accelerated motion with constant acceleration (
towards the ground). Therefore, we can use the following suvat equation:

where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
In this problem, we have:
u = 44.1 m/s is the initial vertical velocity of the ball
v = 0 is the final velocity when the ball reaches the maximum height
s is the maximum height
is the acceleration of gravity (downward, so negative)
Solving for s, we find the maximum height reached by the ball:

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The free-body diagram of an apple falling through the air has weight of the apple pointing downwards and the air-resistance on the apple acting upwards.
When an object falls from up to the ground, the object falls under in the influence of acceleration due to gravity.
The vertical component of the force on the apple as it falls trough the air is given as;
∑Fy = 0
Fₙ - W = 0
Fₙ = W
where;
- <em>Fₙ is the frictional force on the apple acting upwards</em>
- <em>W is the weight of the apple acting downwards</em>
The free-body diagram of the apple is represented as follows;
↑ Fₙ
Ο
↓ W
Thus, the free-body diagram of an apple falling through the air has weight of the apple pointing downwards and the air-resistance on the apple acting upwards.
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Answer:
v = 0.489 m/s
Explanation:
It is given that,
Mass of a box, m = 1.5 kg
The compression in the spring, x = 6.5 cm = 0.065 m
Let the spring constant of the spring is 85 N/m
We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.


So, the speed of the box is 0.489 m/s.
1) Equivalent resistance, 1/R = 1/15 + 1/40 + 1/60 = 8+3+2 /120 = 13/120
R = 120/13 = 9.23
2) Current, I = V/R = 115/9.23 = 12.45 A