If a ball is thrown directly upwards with twice the initial speed of another, how much
higher will it be at its apex?
A) 8 times D) 2 times
B) 4 times E) 2 times
C) 2 2 times
Please show any equations that are used :)
Well, to answer this question, let's assume that the initial speed of the second ball (the one with half the initial speed) is v. Therefore, the initial speed of the first ball (the one thrown directly upwards with twice the initial speed) would be 2v.
Now, let's consider the equation for calculating the maximum height (h) reached by an object thrown vertically upwards:
h = (v^2) / (2g)
Where:
- v is the initial velocity
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
So, for the first ball (with initial speed 2v), the maximum height reached would be:
h1 = ((2v)^2) / (2g)
= (4v^2) / (2g)
= 2(v^2) / (g)
For the second ball (with initial speed v), the maximum height reached would be:
h2 = (v^2) / (2g)
Now, let's find the ratio between h1 and h2:
h1 / h2 = (2(v^2) / (g)) / ((v^2) / (2g))
= (2(v^2) / (g)) * ((2g) / (v^2))
= 4
Therefore, the first ball will be 4 times higher at its apex than the second ball. So, the answer is:
B) 4 times
Hope that clears things up! If not, I can always become a clown and entertain you with some juggling while you figure it out! 🤡
To solve this problem, let's use the equations of motion for projectile motion.
The equation for the maximum height reached by a ball thrown upwards (in the absence of air resistance) is given by:
h = (v^2) / (2g)
Where:
h = maximum height reached
v = initial vertical velocity
g = acceleration due to gravity (approximately 9.8 m/s^2)
Let's assume the initial velocity of the first ball is V, and the initial velocity of the second ball is 2V.
For the first ball:
h1 = (V^2) / (2g)
For the second ball:
h2 = ((2V)^2) / (2g)
= (4V^2) / (2g)
= (V^2) / (0.5g)
= (V^2) / (2g/4)
= (V^2) / (h1/4)
= 4h1
Therefore, the second ball will be 4 times higher at its apex compared to the first ball.
Answer: B) 4 times
To determine the height difference between the two balls, we can use the kinematic equation for vertical displacement. The equation is:
Δy = V₀y * t + (1/2) * a * t²
Where:
Δy is the vertical displacement (height)
V₀y is the initial vertical velocity
t is the time
a is the acceleration (gravitational acceleration, which is approximately -9.8 m/s²)
Let's assume that the initial velocity of the first ball (ball A) is V₀ and the initial velocity of the second ball (ball B) is 2V₀ (since it is twice the speed of ball A).
For both balls, at the apex (highest point), the vertical displacement (Δy) will be zero. Thus, we can set up the equation for both balls as follows:
For ball A:
0 = V₀ * t + (1/2) * (-9.8) * t²
For ball B:
0 = 2V₀ * t + (1/2) * (-9.8) * t²
We can solve the equations to find the value of t for each ball.
For ball A:
0 = V₀ * t + (1/2) * (-9.8) * t²
0 = V₀ * t - 4.9 * t²
For ball B:
0 = 2V₀ * t + (1/2) * (-9.8) * t²
0 = 2V₀ * t - 4.9 * t²
Now, we can divide the two equations to eliminate V₀ and solve for t:
(V₀ * t - 4.9 * t²) / (2V₀ * t - 4.9 * t²) = 0
Simplifying the equation gives us:
t / 2t = 1 / 2
So, t = 2t.
This means that the time taken to reach the apex for both balls is the same.
Now, let's calculate the height at the apex for both balls using the equation:
Δy = V₀y * t + (1/2) * a * t²
For ball A:
Δy = V₀ * t + (1/2) * (-9.8) * t²
Δy = V₀ * t - 4.9 * t²
For ball B:
Δy = 2V₀ * t + (1/2) * (-9.8) * t²
Δy = 2V₀ * t - 4.9 * t²
Since the time taken (t) is the same for both balls, we can compare the coefficients of t² to determine the height difference.
Comparing the two equations:
Ball A: V₀ * t - 4.9 * t²
Ball B: 2V₀ * t - 4.9 * t²
We can see that the coefficient of t is 2 times greater for ball B than ball A. Therefore, the height difference at the apex will be 2 times higher for ball B compared to ball A.
So, the answer is E) 2 times.
You don't need an equation to answer this. The ball thrown at twice the velocity has four times the kinetic energy of the slower ball (assumiung that they have the same mass). The distance that the ball rises is proportional to the initial kinetic energy. So it rises four times as high.
The same answer applies even if the masses are different. That is because
H(max) = Vo^2/(2*g)
Vo is the initial velocity