A ball is tossed vertically upward from a height of 2 m with an initial velocity
of 10 m/s. Just before the ball hits the ground, its velocity will be what?
Vf^2 = Vo^2 + 2gd,
d = ((Vf^2-Vo^2) / 2g) + 2,
d=((0-100) / -19.6) + 2=5.1 + 2 =7.1m
= Distance from Gnd.
Vf^2 = Vo^2 + 2gd,
Vf^2 = 0 + 2*9.8*7.1 = 139.16,
Vf = 11.8m/s.
Well, if the ball is about to hit the ground, I hope it's prepared for a crash landing! But let me calculate its velocity just before that happens.
To do that, we need to use some physics. The motion of the ball can be described by the equation:
v^2 = u^2 + 2as
where:
v is the final velocity (what we're trying to find),
u is the initial velocity (10 m/s in this case),
a is the acceleration (which is the acceleration due to gravity, approximately -9.8 m/s^2),
and s is the distance traveled (which is 2 m in this case).
Plugging in these values, we can solve for v:
v^2 = (10 m/s)^2 + 2*(-9.8 m/s^2)*(2 m)
v^2 = 100 m^2/s^2 + 2*(-9.8 m/s^2)*(2 m)
v^2 = 100 m^2/s^2 - 39.2 m^2/s^2
v^2 = 60.8 m^2/s^2
Taking the square root of both sides, we get:
v = √(60.8 m^2/s^2)
v ≈ 7.8 m/s
So, just before the ball smacks into the ground, its velocity will be approximately 7.8 m/s. It's gonna bounce back like a jumpy bean!
To determine the velocity just before the ball hits the ground, we need to consider the motion of the ball.
When a ball is tossed vertically upward, it experiences a constant force of gravity pulling it downward. As a result, its velocity decreases until it reaches the highest point of its trajectory and then begins to fall back down.
We can use the kinematic equation to calculate the final velocity of the ball just before hitting the ground:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration (due to gravity, which is approximately -9.8 m/s^2)
s = displacement (in this case, the distance fallen by the ball)
Given:
u = 10 m/s (initial velocity)
s = the total distance fallen by the ball
We can find s by calculating the total height from which the ball was thrown (2 m) plus the total distance fallen:
s = 2 m + (2 m × 2)
s = 2 m + 4 m
s = 6 m
Now, we can plug these values into the equation:
v^2 = 10^2 + 2(-9.8)(6)
v^2 = 100 + (-117.6)
v^2 = -17.6
Since velocity cannot be negative in this context, we can take the square root of both sides of the equation:
v = √(-17.6)
The square root of a negative number is not a real number, which implies that the ball will not hit the ground again. Therefore, just before the ball hits the ground, its velocity will be zero.
To find the velocity of the ball just before it hits the ground, we can use the principles of physics. The key concept to apply in this scenario is the conservation of energy.
1. Start by calculating the initial potential energy of the ball when it is at a height of 2m. The formula for potential energy is PE = m * g * h, where m is the mass of the ball (which we can assume to be constant for simplicity), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height (2m).
PE = m * g * h
PE = m * 9.8 * 2
PE = 19.6 * m Joules
2. Next, calculate the initial kinetic energy of the ball using the formula KE = 0.5 * m * v^2, where v is the initial velocity of the ball (10 m/s).
KE = 0.5 * m * v^2
KE = 0.5 * m * 10^2
KE = 50 * m Joules
3. Since energy is conserved, the initial potential energy is equal to the initial kinetic energy when the ball is at its maximum height. Therefore, we equate the two expressions and solve for the initial velocity (v).
PE = KE
19.6 * m = 50 * m
19.6 = 50
Solving for m, we get m = 0.392 kg.
4. Now, we know the mass of the ball, m, we can calculate its velocity just before it hits the ground. Using the equation v^2 = u^2 + 2as, where u is the initial velocity, a is the acceleration due to gravity (-9.8 m/s^2 as it opposes the upward motion), and s is the distance (2m), we can solve for v.
v^2 = u^2 + 2as
v^2 = 10^2 + 2 * (-9.8) * 2
v^2 = 100 + (-39.2)
v^2 = 60.8
v ≈ √60.8
v ≈ 7.8 m/s
Therefore, just before the ball hits the ground, its velocity will be approximately 7.8 m/s (rounded to one decimal place).