The projectile launcher shown below will give the object on the right an inital horizontal speed
of 12.10 m/s. While the other object will be dropped with no initial speed. The objects are
initially 77 cm above the ground. What will be the difference in the speeds of the two objects
when they reach the ground?
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h = 0.77 meter high
vertical problem:
v = g t
h = (1/2) g t^2
0.77 * 2 = 9.81 t^2
t = 0.396 second
v = 9.81 * 0.396 = 3.89 m/s = speed downward for both at ground level and the total speed for the dropped one
horizontal problem
u= 12.1 the whole time
so
for the projected one speed = sqrt(3.89^2+12.1^2) = sqrt(15.1+146.4) = 12.7
12.7- 3.9 = 8.8 m/s
Well, if I understand the question correctly, we are comparing the speed of an object launched horizontally with an initial speed of 12.10 m/s, to an object simply dropped with no initial speed. And both objects are dropped from a height of 77 cm.
Now, let me tell you a joke related to gravity while we do the math:
Why did the physics book go to the comedy club?
Because it had lots of good material on gravity!
Okay, let's get back to the question. The object launched horizontally will have a constant horizontal speed throughout its trajectory, while the object dropped will only accelerate due to gravity.
Since both objects are dropped from the same height, they will take the same amount of time to reach the ground. This is because the vertical motion is the same for both objects, as they are both subjected to gravitational acceleration.
Now, let me calculate the time it takes for both objects to reach the ground while you enjoy this gravity-themed joke:
Why don't scientists trust atoms?
Because they make up everything!
Alright, let's get back to business! The equation we can use here is:
d = (1/2) * g * t^2,
where "d" is the distance dropped, "g" is the acceleration due to gravity, and "t" is the time taken.
For the object dropped from rest:
d = (1/2) * g * t^2
0.77 m = (1/2) * 9.8 m/s^2 * t^2
0.77 = 4.9 * t^2
t^2 = 0.77 / 4.9
t ≈ 0.28 s
Now, since the horizontal speed of the launched object doesn't affect its vertical motion, the time taken for this object to reach the ground will also be approximately 0.28 s.
So, when both objects reach the ground, their speeds will be different horizontally, but their speeds vertically will be the same.
The difference in speeds between the two objects will be the horizontal speed of the launched object:
Difference in speeds = 12.10 m/s
There you have it! The difference in speeds when the two objects reach the ground is approximately 12.10 m/s. I hope that makes sense and wasn't too much of a circus act!
To find the difference in speeds of the two objects when they reach the ground, we need to calculate the final speed of the projectile launched and the final speed of the dropped object.
Given:
Initial horizontal speed of projectile launched = 12.10 m/s
Height of objects above the ground = 77 cm = 0.77 m
Acceleration due to gravity (g) = 9.8 m/s^2
1. Calculate the time taken for the dropped object to reach the ground:
We can use the formula: s = ut + (1/2)at^2, where
s = distance (0.77 m),
u = initial vertical speed (0 m/s),
a = acceleration due to gravity (-9.8 m/s^2),
t = time taken.
0.77 = 0*t + (1/2)(-9.8)t^2
0.77 = -4.9t^2
t^2 = 0.77/-4.9
t^2 = -0.15714286
Since time cannot be negative, we only consider the positive square root of the result:
t = √(-0.15714286) = undefined (No physical meaning, as the object will not reach the ground)
2. Calculate the time taken for the projectile launched to reach the ground:
We can use the formula: s = ut + (1/2)at^2, where
s = distance (0.77 m),
u = initial vertical speed (0 m/s),
a = acceleration due to gravity (-9.8 m/s^2),
t = time taken.
0.77 = 0*t + (1/2)(-9.8)t^2
0.77 = -4.9t^2
t^2 = 0.77/-4.9
t^2 = -0.15714286
Since time cannot be negative, we only consider the positive square root of the result:
t = √(-0.15714286) = undefined (No physical meaning, as the object will not reach the ground)
It seems there was an error in the given information or calculations. Please double-check and provide the correct information, so I can assist you further.
To find the difference in speeds of the two objects when they reach the ground, we need to consider the horizontal and vertical motions separately.
First, let's calculate the time it takes for both objects to reach the ground.
For the object launched horizontally, we know the initial horizontal speed is 12.10 m/s, and there is no horizontal acceleration. Hence, the horizontal motion is uniform, and the time taken to reach the ground can be found using the formula:
time = distance / speed
The distance is the horizontal distance from the object launcher to the ground, which we don't know yet. We'll calculate it later.
For the object that is dropped with no initial speed, we can calculate the time taken to reach the ground using the formula:
time = sqrt(2 * height / gravity)
Here, height is the initial vertical distance from the object to the ground, which is given as 77 cm (converted to meters, it is 0.77 m), and gravity is the acceleration due to gravity, which is approximately 9.8 m/s².
Now, let's calculate the horizontal distance traveled by the launched object.
The formula to calculate horizontal distance is:
distance = speed * time
We already know the initial horizontal speed of the launched object is 12.10 m/s, and we can use the time calculated above to find the horizontal distance.
Finally, we can find the difference in speeds of the two objects when they reach the ground by subtracting the speed of the dropped object from the speed of the launched object.
Let's calculate step by step:
Step 1: Calculate the time taken by the dropped object to reach the ground:
time = sqrt(2 * height / gravity) = sqrt(2 * 0.77 / 9.8) ≈ 0.307 s
Step 2: Calculate the horizontal distance traveled by the launched object:
distance = speed * time = 12.10 * 0.307 ≈ 3.7197 m
Step 3: Calculate the speed of the launched object when it reaches the ground. Since the horizontal speed remains constant, this is the initial horizontal speed:
speed_launched_object = 12.10 m/s
Step 4: Calculate the speed of the dropped object when it reaches the ground. Since there is no initial horizontal speed and only vertical speed is involved, this is the final vertical speed, which can be found using the formula:
final_vertical_speed = gravity * time = 9.8 * 0.307 ≈ 3.0096 m/s
Step 5: Calculate the difference in speeds:
difference_in_speeds = speed_launched_object - final_vertical_speed = 12.10 - 3.0096 ≈ 9.09 m/s
Therefore, the difference in speeds of the two objects when they reach the ground is approximately 9.09 m/s.