a man throws his keys up to his wife 4.00m above. the wife catches the keys 1.50s later. with what initial velocity were the keys thrown?
hf=hi+vi*t-4.9t^2
4=0+vi*1.5 - 4.9*(1.5)^2
solve for vi
10.017m/s
To find the initial velocity with which the man threw the keys, we can use the kinematic equation:
Distance (d) = (Initial Velocity (u) * Time (t)) + (0.5 * Acceleration (a) * Time (t)^2)
Let's break down the given information:
Distance (d) = 4.00 m
Time (t) = 1.50 s
Acceleration (a) = -9.81 m/s^2 (assuming constant acceleration due to gravity)
We need to rearrange the equation to solve for the initial velocity (u):
d = ut + 0.5at^2
4.00 = u * 1.50 + 0.5*(-9.81) * (1.50^2)
Now we can solve for the initial velocity (u):
4.00 = 1.5u - 10.58325
1.5u = 14.58325
u = 9.72217 m/s
Therefore, the man threw the keys with an initial velocity of approximately 9.72 m/s.