To determine the time it takes for the mass to return to the ground and the velocity at which it hits the ground, we need to consider the motion of the mass under the influence of gravity.
When an object is thrown vertically upward, it will experience a deceleration due to the force of gravity until it reaches its maximum height. At the highest point, the velocity will be momentarily zero, and then it will start descending back towards the ground, accelerating due to gravity.
Given: Initial velocity (u) = 10 m/s (upward) Final velocity (v) = ? (at the ground) Acceleration due to gravity (g) = 9.8 m/s^2 (assuming standard Earth gravity)
- To find the time it takes for the mass to return to the ground, we can use the fact that the total displacement (change in height) will be zero when it reaches the ground. We can use the equation:
v^2 = u^2 + 2as
Where: v = final velocity (0 m/s at the highest point) u = initial velocity (10 m/s) a = acceleration due to gravity (-9.8 m/s^2, taking downward as negative) s = displacement (height reached, considering upward as positive and downward as negative)
0^2 = 10^2 + 2*(-9.8)*s
Solving for s: -98s = -100 s = 100/98 ≈ 1.02 m
The total displacement (height) is 1.02 m.
Next, we can calculate the time it takes to reach that height: v = u + at
0 = 10 + (-9.8)*t
Solving for t: -9.8t = -10 t = 10/9.8 ≈ 1.02 s
The time it takes for the mass to reach its maximum height is approximately 1.02 seconds.
- Finally, to determine the velocity at which it hits the ground, we can use the equation:
v = u + at
v = 10 + (-9.8)*t
Substituting the time t = 1.02 s:
v = 10 + (-9.8)*(1.02) v ≈ 0.196 m/s (rounded to three decimal places)
The velocity at which the mass hits the ground is approximately 0.196 m/s (downward).
Note: The negative sign indicates the downward direction.