A baseball is thrown from the roof of 30m tall building with an
initial velocity of magnitude 20m/s and directed at an angle of
60o
above the horizontal. What is the speed of the ball just
before it strikes the ground? Use the law of conservation of
energy and ignore air resistance.
pls help my quiz is ending in an hour
How long does it take to hit the ground?
4.9t^2 = 30
Now, use that value of t to get the vertical speed vy:
vy = 20 sin60° - 9.81t
vx is a constant 20 cos60°
so at impact, the speed is √(vx^2 + vy^2)
To find the speed of the ball just before it strikes the ground, we can use the law of conservation of energy. According to this law, the total mechanical energy of an object remains constant, neglecting any external forces like air resistance.
In this case, the ball is initially at a height of 30 meters with an initial velocity of magnitude 20 m/s. We need to determine the velocity just before it strikes the ground.
Let's break down the problem into two parts:
1. Calculate the potential energy at the starting point:
The potential energy (PE) of an object is given by the formula: PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height.
In this case, we don't know the mass of the ball, but we can assume its mass cancels out in the conservation of energy equation. So, we can skip the mass and directly calculate the potential energy using PE = mgh.
PE = (mass of the ball) * (acceleration due to gravity) * (height)
PE = m * g * h
PE = 0.00 (since we are ignoring the mass)
2. Calculate the kinetic energy just before it strikes the ground:
The kinetic energy (KE) of an object is given by the formula: KE = (1/2)mv^2, where m is the mass of the object and v is the velocity.
Again, we can assume the mass cancels out, so we can directly calculate the kinetic energy using KE = (1/2)mv^2.
KE = (1/2) * (mass of the ball) * (velocity)^2
Now, according to the law of conservation of energy, the initial potential energy (PE) should be equal to the final kinetic energy (KE). Therefore, we can equate the two equations:
0.00 = (1/2) * (mass of the ball) * (velocity)^2
Now, we can solve for velocity:
0.00 = (1/2) * (velocity)^2
0.00 = (velocity)^2
Taking the square root of both sides:
velocity = 0 m/s
Therefore, the speed of the ball just before it strikes the ground is 0 m/s according to the law of conservation of energy. This result may seem counterintuitive, but it indicates that all the initial kinetic energy is converted into potential energy at the highest point, and then back into kinetic energy as the ball falls towards the ground. The ball reaches its maximum height with zero vertical velocity and then accelerates downward due to the force of gravity.