A projectile is launched from and returns to ground level, as the figure shows. There is no air resistance. The horizontal range of the projectile is measured to be R= 166 m, and the horizontal component of the launch velocity is v0x = 26.0 m/s. Find the vertical component v0y of the projectile.
time in air = 166/26 = 6.38 s
so what stays up that long?
well, it spends half that time going up
t = 3.19 s upward
v = Vi - g t
at top v = 0
0 = Vi - 9.81(3.19)
Vi = 31.3 m/s initial speed up
Well, you've got a projectile going up and down like a yo-yo. And you want to find the vertical component of its launch velocity? Alrighty then!
Since there's no air resistance, we can use some good ol' physics to figure this out.
First, let's break down the motion into its horizontal and vertical components. The horizontal component doesn't change throughout the projectile's journey because nothing is acting upon it in that direction. So we'll just stick with the given value of v0x = 26.0 m/s.
Now, when the projectile is launched, it goes up, reaches its maximum height, and then comes back down to the ground. At its highest point, the vertical component of its velocity is 0 m/s because it momentarily stops before changing direction.
We can use this information to find the time it takes for the projectile to go from the ground to its peak. And since the vertical velocity is 0 m/s at its peak, we can use the following equation:
v0y - g * t = 0
Where v0y is the initial vertical component of the velocity and g is the acceleration due to gravity (-9.8 m/s^2).
Solving for t, we get:
t = v0y / g
Now, we know that the total time it takes for the projectile to go up and come back down is twice the time it takes to reach the peak. So:
2 * t = total time of flight
We can calculate the total time using the horizontal range, R, and the horizontal component of the velocity, v0x. The formula for range is:
R = v0x * total time
Rearranging the equation, we have:
total time = R / v0x
Plugging in the given values, we get:
total time = 166 m / 26.0 m/s = 6.385 s
Since the total time is twice the time it takes to reach the peak, we can now find the time it takes for the projectile to reach the peak by dividing the total time by 2:
t = total time / 2 = 6.385 s / 2 = 3.1925 s
Now that we have the time it takes to reach the peak, we can go back to the equation we had before:
t = v0y / g
Plugging in the values, we have:
3.1925 s = v0y / -9.8 m/s^2
Solving for v0y:
v0y = -9.8 m/s^2 * 3.1925 s = -31.2135 m/s
And remember, the negative sign indicates that the velocity is in the opposite direction of the gravitational force.
So, the vertical component of the launch velocity of the projectile is approximately -31.21 m/s.
Hope that helps! And remember, don't let gravity bring you down! Keep reaching for the sky!
To find the vertical component v0y of the projectile, we can use the equation for the horizontal range of a projectile:
R = (v0x * t)
where R is the horizontal range, v0x is the horizontal component of the launch velocity, and t is the time of flight.
Since the projectile launches from and returns to ground level, we know that the time of flight will be twice the time it takes for the projectile to reach its maximum height. Let's call this time t_max.
We can use the equations of motion to find t_max. The vertical motion of the projectile can be described by the equation:
y = v0y * t - (1/2) * g * t^2
where y is the vertical distance, v0y is the vertical component of the launch velocity, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
At the maximum height, the vertical distance y is zero. Therefore, we can set the equation equal to zero:
0 = v0y * t_max - (1/2) * g * t_max^2
Rearranging the equation, we get:
t_max * (v0y - (1/2) * g * t_max) = 0
This equation has two solutions: t_max = 0 or v0y - (1/2) * g * t_max = 0
Since time cannot be zero in this case, we can equate the second part of the equation to zero:
v0y - (1/2) * g * t_max = 0
Solving for v0y, we get:
v0y = (1/2) * g * t_max
Now, we need to find t_max in terms of R and v0x. The time t for the projectile to reach the maximum height can be found using the equation:
v0y = v0 - g * t
At the maximum height, the vertical component of the velocity is zero:
0 = v0y - g * t
Simplifying this equation, we get:
t = v0y / g
Substituting this value for t in terms of v0y into the equation for R, we get:
R = v0x * (2 * t)
Substituting v0y / g for t, we have:
R = v0x * (2 * v0y / g)
Simplifying further, we get:
R = (2 * v0x * v0y) / g
Rearranging this equation, we can solve for v0y:
v0y = (R * g) / (2 * v0x)
Using the given values, with R = 166 m and v0x = 26.0 m/s, and substituting the acceleration due to gravity g = 9.8 m/s^2, we can calculate v0y:
v0y = (166 * 9.8) / (2 * 26.0)
v0y = 8.825 m/s
Therefore, the vertical component of the projectile's velocity is v0y = 8.825 m/s.
To find the vertical component v0y of the projectile, we can use the equation of motion and the given information.
The equation of motion for vertical motion can be written as:
y = v0y * t - (1/2) * g * t^2
where:
- y is the vertical displacement
- v0y is the initial vertical velocity
- t is the time
- g is the acceleration due to gravity (-9.8 m/s^2)
In this case, the projectile is launched and returns to ground level, which means the vertical displacement is zero. So we can rewrite the equation as:
0 = v0y * t - (1/2) * g * t^2
The time t can be obtained using the horizontal range R and the horizontal component of the launch velocity v0x. The time of flight (time taken to travel horizontally) is given by:
t = R / v0x
Substituting this value of t into the equation, we get:
0 = v0y * (R / v0x) - (1/2) * g * (R / v0x)^2
Simplifying this equation, we can solve for v0y:
v0y = (1/2) * g * (R / v0x)
Now, let's substitute the given values into the equation:
v0y = (1/2) * (9.8 m/s^2) * (166 m / 26.0 m/s)
Calculating this, we can find the value of v0y.