a ball thrown 45m high speed of 10m/s at what time reach the top?
To find the time it takes for the ball to reach the top, we can use the kinematic equation for vertical motion:
\(v_f = v_i + at\)
Where:
\(v_f\) = final velocity (which is 0 m/s at the top)
\(v_i\) = initial velocity (10 m/s, upwards)
\(a\) = acceleration (due to gravity, which is approximately -9.8 m/s²)
\(t\) = time
Since the ball reaches the top when its final velocity becomes 0 m/s, we can substitute the known values into the equation to solve for \(t\):
\(0 = 10 - 9.8t\)
Simplifying the equation gives us:
\(9.8t = 10\)
Now we can solve for \(t\):
\(t = \frac{10}{9.8} \approx 1.02\) seconds
Therefore, it takes approximately 1.02 seconds for the ball to reach the top.