A ball is thrown vertically upward with a velocities of 18 m/s. Two seconds later another ball is thrown upwards with a velocity of 13.5 m/s. At what position above the ground will they meet?
you want to find where
18t - 4.9t^2 = 13.5(t-2) - 4.9(t-2)^2
t = 3.086 seconds
height: 8.883 m
To find the position at which the two balls meet, we need to consider their respective heights as a function of time.
Let's start by finding the equation for the first ball's height as a function of time. When a ball is thrown vertically upward, it experiences a constant downward acceleration due to gravity (9.8 m/s²). Using the kinematic equation for displacement, we have:
s₁ = v₁t - (1/2)gt²
where:
s₁ = height of the first ball (which we're looking for)
v₁ = initial velocity of the first ball (18 m/s)
t = time
Now, let's find the equation for the second ball's height as a function of time. The second ball is thrown 2 seconds later, so we need to account for the time difference. Using the same kinematic equation, we have:
s₂ = v₂(t - 2) - (1/2)g(t - 2)²
where:
s₂ = height of the second ball (which we're looking for)
v₂ = initial velocity of the second ball (13.5 m/s)
t = time
Now, to find the position where the two balls meet, we need to set the heights equal to each other and solve for the time (t):
s₁ = s₂
v₁t - (1/2)gt² = v₂(t - 2) - (1/2)g(t - 2)²
Now we can rearrange and solve for t:
v₁t - v₂(t - 2) = (1/2)g(t - 2)² - (1/2)gt²
v₁t - v₂t + 2v₂ = (1/2)gt² - g(t - 2)²
v₁t - v₂t + 2v₂ = (1/2)gt² - g(t² - 4t + 4)
v₁t - v₂t + 2v₂ = (1/2)(-gt² + 4gt - 4g)
v₁t - v₂t + 2v₂ = -1/2gt² + 2gt - 2g
Now, we can simplify the equation further:
1/2gt² + v₁t - v₂t + 2v₂ = 2gt - 2g
1/2gt² + t(v₁ - v₂) + 2v₂ = 2gt - 2g
1/2gt² + t(v₁ - v₂ - 2g) + 2v₂ + 2g = 0
Finally, we can solve this quadratic equation for t. Once we find the value of t, we can substitute it back into either s₁ or s₂ equation to find the position where the two balls meet.