A stone is thrown up from the top of a platform with an initial velocity of 19.8m/s. If the top of the platform is taken to be the ground level, calculate The Velocity just before reaching the ground
it will be the same as when it went up.
Since the platforms are both of the same height, the velocities will be exactly the same, albeit in a negative direction.
Therefore, your answer is -19.8mls.
Well, the stone seems to be having quite a "rocky" journey! Let's calculate its velocity just before it reaches the ground.
When the stone is thrown up, it experiences a downward gravitational force that slows it down. At the highest point of its trajectory, its velocity is 0 m/s. Using this information, we can figure out the time it takes for the stone to reach the highest point.
Since we know the initial velocity (u), the acceleration (a) due to gravity (-9.8 m/s^2), and the final velocity (v) at the highest point, we can use the equation v = u + at and solve for t.
0 = 19.8 m/s + (-9.8 m/s^2)t
Solving this equation, we find t = 2 seconds. Righto!
Now, let's find out the velocity just before it reaches the ground. Since the time taken to go up is 2 seconds, it will also take 2 seconds to come back down.
Using the equation v = u + at, where t = 2 seconds, u = 19.8 m/s, and a = 9.8 m/s^2 (since the stone is now moving downwards), we can calculate the final velocity (v) just before it reaches the ground.
v = 19.8 m/s + (9.8 m/s^2)(2 seconds)
v = 19.8 m/s + 19.6 m/s
v = 39.4 m/s
So, the velocity just before the stone reaches the ground is approximately 39.4 m/s. Phew! That's quite a "speedy" descent!
To calculate the velocity of the stone just before reaching the ground, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement
In this case, we know the initial velocity (u = 19.8 m/s) and the acceleration due to gravity (a = -9.8 m/s^2). The negative sign is used because acceleration due to gravity acts in the opposite direction to the initial velocity.
The displacement (s) of the stone when it reaches the ground is equal to the height of the platform. Since the platform is at the ground level, the displacement is zero.
Plugging in the known values into the kinematic equation, we have:
v^2 = (19.8 m/s)^2 + 2 (-9.8 m/s^2)(0)
v^2 = 392.04 m^2/s^2
Therefore, the velocity of the stone just before reaching the ground is the square root of 392.04 m^2/s^2, which is approximately 19.8 m/s (rounded to two decimal places).