a boy throws a stone vertically up to a man standing at a height of 2.0m above the boy. If the stone is thrown up with a velocity of 7.0metres per second what is the velocity of the stone at the instant when it is caught by the man
A. 3.0m/s B. 6.2m/s C. 9.0m/s D. 9.4m/s
m g h = (1/2)m (7^2) - (1/2) m v^2
9.81 (4) = 49 - v^2
v^2 = about 9
v = about 3
Good
To solve this problem, we can use the principle of conservation of energy. At the instant when the stone is caught by the man, its kinetic energy will be converted to potential energy.
The initial kinetic energy of the stone is given by:
KE_initial = 1/2 * m * v_initial^2
where m is the mass of the stone and v_initial is its initial velocity.
The final potential energy of the stone when caught by the man is given by:
PE_final = m * g * h
where g is the acceleration due to gravity and h is the height of the man.
Since the stone is caught at the same height from which it was thrown, the final potential energy is equal to the initial kinetic energy:
PE_final = KE_initial
Now, we can set up the equation and solve for the final velocity.
1/2 * m * v_initial^2 = m * g * h
We can simplify the equation by canceling out the mass:
1/2 * v_initial^2 = g * h
Substituting in the given values:
1/2 * (7.0 m/s)^2 = 9.8 m/s^2 * 2.0 m
Simplifying further:
1/2 * 49 m^2/s^2 = 19.6 m^2/s^2
Therefore, the final velocity of the stone when caught by the man is 7.0 m/s.
So, the answer is not in the given choices.
To find the velocity of the stone at the instant when it is caught by the man, we can use the equation of motion for vertical motion:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
The stone is thrown vertically upwards, so we can consider its motion to be free fall under gravity. The acceleration due to gravity is approximately 9.8 m/s², but since the stone is thrown upwards, the acceleration will be in the opposite direction.
Given:
Initial velocity (u) = 7.0 m/s (upwards)
Final velocity (v) = ?
Acceleration (a) = -9.8 m/s² (downwards, as it opposes the stone's upward motion)
Time (t) = ?
We also know that the stone is caught by the man at a height of 2.0 m above the boy. We can use this information to calculate the time it takes for the stone to reach the man.
Using the equation of motion for vertical displacement:
s = ut + 0.5at²
where:
s = displacement
u = initial velocity
t = time
a = acceleration
Rearranging the equation, we get:
t = √(2s/a)
Substituting the values:
s = 2.0 m
a = -9.8 m/s²
t = √(2(2.0)/-9.8)
t = √(4/-9.8)
t ≈ 0.64 s
Now that we know the time it takes for the stone to reach the man, we can find the final velocity (v) using the equation of motion:
v = u + at
v = 7.0 + (-9.8)(0.64)
v = 7.0 - 6.27
v ≈ 0.73 m/s
Therefore, the velocity of the stone at the instant when it is caught by the man is approximately 0.73 m/s.
None of the provided answer options (A: 3.0 m/s, B: 6.2 m/s, C: 9.0 m/s, D: 9.4 m/s) match the calculated velocity.