The projectile launcher shown below will give the object on the right an inital horizontal speed of 9.40 m/s. While the other object will be dropped with no initial speed. The objects are initially 114 cm above the ground. What will be the difference in the speeds of the two objects when they reach the ground?
I will have to guess at your drawing.
Both objects fall the same.
The first one keeps its horizontal velocity
so
both fall 1.14 meters
v = g t = 9.81 t
1.14 = (1/2) g t^2 = 4.9 t^2
so t = 0.482 second fall time for both
so vertical velocity of both is
9.81 * .482 = 4.73 m/s
that is the entire velocity of the dropped one
for the projectile
speed = sqrt(4.73^2 + 9.4^2)
so calculate that and subtract 4.73 to get the speed difference
To find the difference in speeds of the two objects when they reach the ground, we need to calculate the vertical speed (or velocity) of each object separately.
1. For the dropped object:
Since it is dropped, it doesn't have an initial vertical speed. The vertical speed will be solely due to the acceleration due to gravity (which is approximately 9.8 m/s^2).
Using the formula for motion under constant acceleration:
v = u + at
where v is the final vertical speed, u is the initial vertical speed, a is the acceleration, and t is the time taken.
In this case, the initial vertical speed is 0 (as object is dropped without any initial speed) and the acceleration due to gravity is -9.8 m/s^2 (negative due to the downward direction). We need to find the time taken for the object to reach the ground.
Using the formula:
s = ut + (1/2)at^2
where s is the vertical displacement (distance traveled), u is the initial vertical speed, a is the acceleration, and t is the time taken.
In this case, the vertical displacement is 114 cm = 1.14 m, the initial vertical speed u is 0, and the acceleration is -9.8 m/s^2. We can solve this equation for time (t).
1.14 = 0 + (1/2)(-9.8)t^2
1.14 = -(4.9)t^2
t^2 = 1.14 / (-4.9)
t^2 = -0.23265
t ≈ -0.482 s (Ignoring the negative root as time cannot be negative)
Now, we can calculate the final vertical speed (v) of the dropped object using:
v = u + at
v = 0 + (-9.8)(-0.482)
v ≈ 4.724 m/s
2. For the projectile-launched object:
The object is given an initial horizontal speed of 9.40 m/s but no initial vertical speed. Therefore, the horizontal speed will remain constant throughout the motion, while only the vertical motion will be affected by acceleration due to gravity.
Using the same formula:
s = ut + (1/2)at^2
The vertical displacement is again 1.14 m, the initial vertical speed (u) is 0, and the acceleration due to gravity (a) is -9.8 m/s^2. We need to solve this equation for time (t).
1.14 = 0 + (1/2)(-9.8)t^2
1.14 = -(4.9)t^2
t^2 = 1.14 / (-4.9)
t^2 = -0.23265
t ≈ -0.482 s (Ignoring the negative root as time cannot be negative)
The final vertical speed of the projectile-launched object (v) is given by:
v = u + at
v = 0 + (-9.8)(-0.482)
v ≈ 4.724 m/s
Therefore, the difference in speeds of the two objects when they reach the ground will be:
Difference = Speed of projectile-launched object - Speed of dropped object
Difference = 4.724 m/s - 4.724 m/s
Difference = 0 m/s
To find the difference in speeds of the two objects when they reach the ground, we first need to calculate the time it takes for each object to reach the ground. Let's assume that the acceleration due to gravity is 9.8 m/s².
For the object on the right launched with an initial horizontal speed of 9.40 m/s, it will experience the same vertical acceleration due to gravity as the dropped object. However, the horizontal speed does not affect the time it takes to reach the ground.
The formula to calculate the time taken for an object to fall vertically from a given height can be derived using the kinematic equation:
s = ut + (1/2)gt²
Where:
s = vertical distance (114 cm or 1.14 m)
u = initial vertical velocity (0 m/s for the dropped object since it's initially at rest)
g = acceleration due to gravity (9.8 m/s²)
t = time taken
Rearranging the equation to solve for t:
1.14 = 0t + (1/2)(9.8)(t²)
1.14 = 4.9t²
t² = 1.14/4.9
t² ≈ 0.232653
t ≈ √0.232653
t ≈ 0.482 s
Therefore, it will take approximately 0.482 seconds for the dropped object to reach the ground.
Now, let's calculate the horizontal distance covered by the projectile launched at 9.40 m/s during this time. Since the horizontal speed remains constant throughout the motion, we can use the equation:
s = ut
s = (9.40)(0.482)
s ≈ 4.5248 m
Therefore, the projectile launched at 9.40 m/s will cover approximately 4.5248 meters horizontally during the time it takes for the dropped object to reach the ground.
Now we know that both objects will reach the ground at the same time (0.482 seconds). However, the object launched at 9.40 m/s travels an additional 4.5248 meters horizontally compared to the dropped object.
To calculate the difference in speeds when they reach the ground, we need to divide the horizontal distance travelled by the additional time taken:
Speed difference = Horizontal distance / Time
Speed difference = 4.5248 m / 0.482 s
Speed difference ≈ 9.38 m/s
Therefore, the difference in speeds of the two objects when they reach the ground is approximately 9.38 m/s.