At time t = 0, a projectile is launched from ground level. At t = 2.00 s, it is displaced d = 57 m horizontally and h = 73 m vertically above the launch point.
At the instant it reaches its maximum height above ground level, what is its horizontal displacement D from the launch point?
79.49
To find the horizontal displacement D from the launch point at the instant the projectile reaches its maximum height above ground level, we can use the equation of motion for horizontal displacement:
D = Vx * t
Where:
- D is the horizontal displacement
- Vx is the horizontal component of the velocity of the projectile
- t is the time elapsed
Since the horizontal component of the velocity remains constant throughout the motion, we can use the average velocity to find Vx.
The average velocity in the horizontal direction can be calculated using the given displacement and time:
Average velocity = displacement / time
Plugging in the given values:
Average velocity = 57 m / 2.00 s = 28.5 m/s
Since the horizontal velocity remains constant, the horizontal component of the velocity (Vx) is also 28.5 m/s.
Now, we need to find the time it takes for the projectile to reach its maximum height. Since the projectile is launched from ground level, reaches a maximum height, and then falls back to the ground, the time it takes to reach the maximum height is half of the total time of flight.
The total time of flight can be found using the vertical displacement and the equation of motion for vertical displacement:
h = Vy0 * t + (1/2) * g * t^2
Where:
- h is the vertical displacement
- Vy0 is the initial vertical component of the velocity
- t is the time elapsed
- g is acceleration due to gravity (approximately 9.8 m/s^2)
At the maximum height, the vertical displacement is zero. Therefore, the equation can be rearranged as follows:
0 = Vy0 * t_max + (1/2) * g * t_max^2
Simplifying:
(1/2) * g * t_max^2 = -Vy0 * t_max
Dividing both sides by t_max:
(1/2) * g * t_max = -Vy0
Since the vertical component of the velocity is positive when launched and becomes negative at the maximum height, we can rewrite this as:
(1/2) * g * t_max = -Vy0
Simplifying:
g * t_max = -2 * Vy0
Solving for t_max:
t_max = -2 * Vy0 / g
Since the initial vertical velocity (Vy0) is positive when launched, we can substitute the given value for Vy0:
t_max = -2 * (73 m) / (9.8 m/s^2)
Calculating:
t_max ≈ -14.9 s
The negative sign indicates that we are considering the time before reaching the maximum height.
To find the time at maximum height (t_max) in relation to the total time of flight (t), we divide t_max by 2:
t_max = t / 2
Solving for t:
t = t_max * 2
Substituting the value of t_max:
t = -14.9 s * 2
Calculating:
t ≈ -29.8 s
The negative sign again indicates that we are considering the time before reaching the maximum height.
Since t can't be negative in this context, we discard the negative value and consider the positive value. Therefore, the total time of flight is approximately 29.8 s.
Now that we know the horizontal velocity (Vx) and the total time of flight (t), we can calculate the horizontal displacement D at the instant the projectile reaches its maximum height:
D = Vx * t
Substituting the known values:
D = 28.5 m/s * 29.8 s
Calculating:
D ≈ 848.3 m
Therefore, the horizontal displacement D from the launch point at the instant the projectile reaches its maximum height above ground level is approximately 848.3 meters.
To find the horizontal displacement D from the launch point when the projectile reaches its maximum height above ground level, we need to use the kinematic equation for horizontal displacement.
The equation for horizontal displacement is:
D = vt
Where:
D is the horizontal displacement
v is the horizontal velocity
t is the time
In this case, we are given the horizontal displacement at t = 2.00 s, which is d = 57 m. We can use this information to find the horizontal velocity v.
First, let's calculate the horizontal velocity v using the formula:
v = d / t
v = 57 m / 2.00 s
v = 28.5 m/s
Now that we have the horizontal velocity, we can calculate the horizontal displacement D when the projectile reaches its maximum height.
Since the horizontal velocity remains constant throughout the projectile's motion, the horizontal displacement D at the maximum height can be found by multiplying the horizontal velocity by the time it takes to reach the maximum height.
The time it takes to reach the maximum height can be found using the equation for vertical displacement:
h = v0t + (1/2)gt^2
Where:
h is the vertical displacement (in this case, h = 73 m)
v0 is the initial vertical velocity (which is equal to the final vertical velocity at maximum height since the projectile is at its peak and about to descend)
t is the time
g is the acceleration due to gravity (approximately -9.8 m/s^2)
Since the projectile reaches its maximum height, its final vertical velocity at that point is 0 m/s. Therefore, we can rearrange the equation as follows:
0 = v0t + (1/2)gt^2
Simplifying the equation, we get:
(1/2)gt^2 = -v0t
Dividing both sides by t, we get:
(1/2)gt = -v0
Simplifying further, we find the relationship between the time and the initial vertical velocity:
v0 = -(1/2)gt
Now we can substitute this value into the equation for horizontal displacement:
D = v * t
D = (57 m/s) * t
Substituting the relationship we found for the initial vertical velocity, we get:
D = (57 m/s) * t
D = (57 m/s) * (-(1/2)gt)
Now we need to solve for t when the projectile reaches its maximum height. At maximum height, the vertical velocity is 0 m/s. We can use this fact to find the time it takes for the projectile to reach maximum height.
0 m/s = v0 + gt
Since the initial vertical velocity is given by v0 = -(1/2)gt, we can substitute this value to solve for t:
0 = -(1/2)gt + gt
Combining like terms, we find:
0 = (1/2)gt
Since the product of two factors is equal to zero only if at least one of the factors is zero, we have:
1/2 = 0 or gt = 0
Since 1/2 ≠ 0, the only solution is gt = 0. This means that the time it takes for the projectile to reach its maximum height is 0 seconds.
Now we can substitute t = 0 into the equation for horizontal displacement:
D = (57 m/s) * (-(1/2)gt)
D = 0
Therefore, the horizontal displacement D from the launch point at the instant the projectile reaches its maximum height above ground level is 0 m.