A stone is thrown horizontal with a velocity of 40.0 m/s from a top of a building find its velocity and displacement after 3.00 seconds
Guys need your help
the horizontal speed is constant at 40 m/s, so the x di
The vertical position at time t is h - 4.9t^2, where h is the height of the building.
The vertical velocity is of course, -9.8t
So plug in t=3 and crank it out.
I need your help
Science
Sure, I can help you with that.
To find the velocity and displacement of the stone after a given time, we can use the equations of motion. In this case, since the stone is thrown horizontally, we only need to consider the horizontal motion.
First, let's find the horizontal velocity of the stone. The stone is thrown horizontally, so there is no vertical acceleration acting on it. We can use the equation:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Since there is no horizontal acceleration, the initial and final velocities in the horizontal direction are the same. Therefore, the horizontal velocity remains constant at 40.0 m/s.
Next, let's determine the displacement of the stone. The displacement in the horizontal direction can be calculated using the equation:
s = ut + (1/2)at^2
where:
s = displacement
u = initial velocity
t = time
In this case, the initial velocity in the horizontal direction is 40.0 m/s and the time is 3.00 seconds. Since there is no horizontal acceleration, the displacement can be calculated as:
s = (40.0 m/s) * (3.00 s) + (1/2)* 0 * (3.00 s)^2
s = 120.0 m
Therefore, the horizontal velocity of the stone after 3.00 seconds is 40.0 m/s and the horizontal displacement is 120.0 m.