A ball is rolled off a horizontal roof at 6m/sec , after leaving the roof how long it will take to reach a speed of 10 m/sec?
A. 0.20 seconds
B. 0.81 seconds
C. 1.8 seconds
D. 2.5 seconds
the horizontal velocity is 6 m/s
the total velocity is the vector sum of the horizontal and vertical components
using Pythagoras, the vertical velocity must be 8 m/s
... it's a 3-4-5 triangle (6-8-10)
B looks good
To find out how long it will take for the ball to reach a speed of 10 m/sec after leaving the roof, we can use the equation for acceleration:
a = (v - u) / t
Here,
a = acceleration (which is equal to g, the acceleration due to gravity)
v = final velocity (10 m/sec)
u = initial velocity (6 m/sec)
We can rearrange the equation to solve for time (t):
t = (v - u) / a
Since the ball is falling freely, the acceleration due to gravity is approximately 9.8 m/s^2 (-9.8 m/s^2, considering the direction). Now, we can substitute the values into the equation:
t = (10 m/sec - 6 m/sec) / (-9.8 m/s^2)
This simplifies to:
t = 4 m/sec / (-9.8 m/s^2)
t ≈ -0.4082 seconds
However, time cannot be negative in this context. Therefore, we can conclude that the ball will not reach a speed of 10 m/sec after leaving the roof.