If a man leans over a cliff and throws a rock upward at 4.9 m/s. Neglecting air resistance, what is the rock's speed one second later?
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v1 = velocity one second later
vi = initial velocity of rock = 4.9 m/s
t = time in seconds
g = acceleration due to gravity = -9.8 m/s^2
v1 = vi + gt
Plug the numbers in and solve for v1.
14.7
To find the rock's speed one second later, we need to consider its initial speed and the effect of gravity.
Initially, the rock is thrown upward with a speed of 4.9 m/s. However, since the rock is moving against the force of gravity, its speed will gradually decrease.
The acceleration due to gravity on Earth is approximately 9.8 m/s² (ignoring air resistance). This means that the rock's speed will decrease by 9.8 m/s every second.
Considering the initial speed of 4.9 m/s and the acceleration due to gravity, we can calculate the rock's speed one second later as follows:
Rock's speed after one second = Initial speed - (Acceleration due to gravity * Time)
Rock's speed after one second = 4.9 m/s - (9.8 m/s² * 1 s)
Rock's speed after one second = 4.9 m/s - 9.8 m/s
Rock's speed after one second = -4.9 m/s
Therefore, after one second, the rock's speed is -4.9 m/s. The negative sign indicates that the rock is moving downward.