ONE SECOND after ball A is projected upwards with a velocity of 16 m/s, a second ball, B, is thrown vertically downwards at a velocity of 9 m/s from a balcony 30 m above the ground.
1.Calculate how high above the ground ball A will be at the instant the two balls pass each other?
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The distance covered by the 1st ball for 1 second is
hₒ=vₒ₁t - gtₒ²/2 =16•1 -9.8•1²/2 = 16 -4.9 = 11.1 m (above the ground)
Its speed at this height is
v= vₒ₁ - gtₒ=16-9.8•1=6.2 m/s.
h=H-hₒ=30-11.1=18.9 m
Now to balls separated by distance ‘h’ are moving towards each other
h₁=vt- gt²/2,
h₂=vₒ₂t + gt²/2,
h= h₁+h₂=vt- gt²/2+ vₒ₂t + gt²/2 = vt+ vₒ₂t = (v+ vₒ₂)t.
t= h/(v+ vₒ₂)=18.9/(6.2+9) = 18/9/15.2=1.24 s.
h₁=vt- gt²/2 =6.2•1.24 -9.8•1.24²/2= =7.688 -7.534=0.154 m
The height above the ground is eaqual to hₒ+h₁=
= 11.1+0.154 = 11.254 m
To find how high above the ground ball A will be at the instant the two balls pass each other, we need to calculate the time it takes for the two balls to meet.
Let's first calculate the time it takes for ball B to reach the ground:
We can use the equation: d = vt + (1/2)at^2
where d is the distance, v is the initial velocity, t is the time, and a is the acceleration. Since ball B is thrown vertically downwards, the acceleration due to gravity will be -9.8 m/s^2 (negative because it is in the opposite direction of the positive direction).
First, let's calculate the time it takes for ball B to reach the ground:
d = 30 m (since the balcony is 30 m above the ground)
v = 9 m/s (initial velocity, downwards)
a = -9.8 m/s^2 (acceleration due to gravity)
Using the equation, we have:
30 = 9t + (1/2)(-9.8)t^2
Rearranging the equation, we get:
-4.9t^2 + 9t - 30 = 0
Solving this quadratic equation, we find two solutions:
t ≈ -0.71 s and t ≈ 4.13 s
Since time cannot be negative in this scenario, we will ignore the negative solution.
So, it takes approximately t = 4.13 s for ball B to reach the ground.
Now, let's find how high ball A will be at this time:
Using the equation: v = u + at
where v is the final velocity (zero when ball A reaches its maximum height), u is the initial velocity, a is the acceleration, and t is the time.
v = 0 m/s (final velocity)
u = 16 m/s (initial velocity, upwards)
a = -9.8 m/s^2 (acceleration due to gravity)
Using the equation, we have:
0 = 16 - 9.8t
Solving for t, we can find the time it takes for ball A to reach its maximum height:
9.8t = 16
t ≈ 1.63 s
So, it takes approximately t = 1.63 s for ball A to reach its maximum height.
Now, we can find how high above the ground ball A will be at the instant the two balls pass each other:
Using the equation: d = 16t + (1/2)(-9.8)t^2
d = 16(1.63) + (1/2)(-9.8)(1.63)^2
d ≈ 26.03 m
Therefore, ball A will be approximately 26.03 m above the ground at the instant the two balls pass each other.
To solve this problem, we need to first calculate the time it takes for ball B to reach the ground. We can use the equation for the motion of an object in free fall:
y = v₀t + 0.5at²
where,
y = displacement (negative because ball B is moving downwards)
v₀ = initial velocity of ball B (9 m/s)
t = time taken for ball B to reach the ground
a = acceleration due to gravity (-9.8 m/s²)
Plugging in the values, we can solve for t:
-30 = 9t + 0.5(-9.8)t²
Rearranging the equation:
4.9t² + 9t + 30 = 0
We can solve this quadratic equation to find the value of t. Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a
where,
a = 4.9
b = 9
c = 30
Substituting the values:
t = (-9 ± √(9² - 4 * 4.9 * 30)) / (2 * 4.9)
Calculating the value of t, we find two possible solutions for t: t = -6 or t = -1.85. Since time cannot be negative, we discard the negative value. Hence, t = -1.85.
Next, we need to find the height of ball A at the same instant when the two balls pass each other. We can use the equation for motion:
y = v₀t + 0.5at²
where,
y = y-axis displacement (height)
v₀ = initial velocity of ball A (16 m/s)
t = time taken for the balls to pass each other (-1.85 seconds)
a = acceleration due to gravity (-9.8 m/s²)
Plugging in the values, we can solve for y:
y = (16 * -1.85) + 0.5 * (-9.8) * (-1.85)²
Calculating the height, we find y is approximately 45.9 meters.
Therefore, at the instant the two balls pass each other, ball A will be approximately 45.9 meters above the ground.