A ball is dropped from a cliff amd has an acceleration of 9.8 meters per second squared. How long will it take the ball to reach a speed of 24.5 meters per second squared (m/s^2) ?
Drag out the old standby formula
s = 1/2 at²
Oops. we're talking velocity
v = at
Note: velocity is not measured in m/s²
To find the time it takes for the ball to reach a certain speed, we can use the kinematic equation:
v = u + at
where:
v = final velocity (24.5 m/s),
u = initial velocity (0 m/s since the ball is dropped),
a = acceleration (-9.8 m/s², negative due to gravity), and
t = time.
Rearranging the equation to solve for time (t), we have:
t = (v - u) / a
Substituting the given values, we have:
t = (24.5 m/s - 0 m/s) / (-9.8 m/s²)
Calculating the value gives us:
t = -24.5 m/s / -9.8 m/s²
t = 2.5 s
Therefore, it will take the ball 2.5 seconds to reach a speed of 24.5 m/s.
To find the time it takes for the ball to reach a certain speed, we can use the formula for calculating final velocity (v), initial velocity (u), acceleration (a), and time (t):
v = u + at
In this case, the ball is dropped, so the initial velocity (u) is 0 m/s. The acceleration (a) is given as 9.8 m/s^2, and the final velocity (v) is 24.5 m/s. We need to find the time it takes, which is represented by (t).
Plugging in the given values into the formula, we have:
24.5 m/s = 0 m/s + (9.8 m/s^2) * t
Simplifying the equation:
24.5 m/s = 9.8 m/s^2 * t
Dividing both sides of the equation by 9.8 m/s^2, we can isolate t:
t = 24.5 m/s / 9.8 m/s^2
Solving the equation:
t = 2.5 seconds
Therefore, it will take the ball 2.5 seconds to reach a speed of 24.5 m/s.