During that final period of time,
his acceleration is
(9 m/s - 5 m/s) / (4 sec) = 1 m/s² .
Did you have a question to ask ?
<span>The ball clears by 11.79 meters
Let's first determine the horizontal and vertical velocities of the ball.
h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s
v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s
Now determine how many seconds it will take for the ball to get to the goal.
t = 36.0 m / 15.04 m/s = 2.394 s
The height the ball will be at time T is
h = vT - 1/2 A T^2
where
h = height of ball
v = initial vertical velocity
T = time
A = acceleration due to gravity
So plugging into the formula the known values
h = vT - 1/2 A T^2
h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2
h = 42.92 m - 4.9 m/s^2 * 5.731 s^2
h = 42.92 m - 28.0819 m
h = 14.84 m
Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
Answer:
f = 878,080 N
Explanation:
mass of pile driver (m) = 2100 kg
distance of pile driver to steel beam (s) = 5 m
depth of steel driven (d) = 12 cm = 0.12 m
acceleration due to gravity (g0 = 9.8 m/s^{2}
calculate the average force exerted on the pile driver by the beam.
- from work done = force x distance
- work done = change in potential energy of the pile driver
- equating the two equations above we have
force x distance = m x g x (s - d)
f x 0.12 = 2100 x 9.8 x (5- (-0.12))
d = - 0.12 because the steel beam went down at we are taking its
initial position to be an origin point which is 0
f = ( 2100 x 9.8 x (5- (-0.12)) ) ÷ 0.12
f = 878,080 N
Answer:
Using g = 9.8: 1.02 kg, Using g = 10: 1 kg
Explanation:
E = mgh
20 = m(9.8)(3 - 1)
20 = 9.8m(2)
20 = 19.6m
m = 1.02 kg
I'm now assuming you may be using a g constant of 10, thus the close integer result, in which case the mass would be exactly 1 kilogram.
