A projectile is shot at a speed of 30m/s at an angle of 50 degrees above the horizontal. The projectile hits a vertical wall at a height 20m above the horizontal. Where is the wall's location? There are 2 answers.
To determine the location of the wall, we need to consider the horizontal and vertical components of the projectile's motion.
Let's start by finding the time it takes for the projectile to hit the wall. We'll only consider the vertical motion for this calculation.
The vertical motion of the projectile can be described by the equation:
h = v₀ₚₑₚᵣₑᵢt + (1/2)at²
Where:
h is the vertical displacement (20m in this case)
v₀ₚₑₚᵣₑᵢ is the initial vertical velocity (which can be found using v₀ₚₑₚᵣₑᵢ = v₀sin(θ))
t is the time taken
a is the acceleration due to gravity (-9.8 m/s²)
Since the projectile is at its highest point when it hits the wall, the vertical displacement is zero. Therefore, the equation becomes:
0 = v₀ₚₑₚᵣₑᵢt - (1/2)gt²
Rearranging the equation, we can solve for t:
t = (2v₀ₚₑₚᵣₑᵢ)/g
Now, let's calculate t:
v₀ₚₑₚᵣₑᵢ = 30 m/s (initial velocity)
θ = 50 degrees (angle)
g = 9.8 m/s² (acceleration due to gravity)
Using the formula v₀ₚₑₚᵣₑᵢ = v₀sin(θ), we can find:
v₀ₚₑₚᵣₑᵢ = 30 m/s * sin(50 degrees) = 23.09 m/s
Substituting these values into the equation for t:
t = (2 * 23.09 m/s) / 9.8 m/s² = 4.73 s
Now that we know the time taken for the projectile to hit the wall is 4.73 seconds, we can determine the horizontal distance traveled by the projectile.
The horizontal motion can be described by the equation:
x = v₀ₚₑₚᵣₑᵢcos(θ) * t
Where:
x is the horizontal distance
v₀ₚₑₚᵣₑᵢ is the initial horizontal velocity (which can be found using v₀ₚₑₚᵣₑᵢ = v₀cos(θ))
θ is the angle of projection (50 degrees in this case)
t is the time taken (4.73 seconds)
Using the formula v₀ₚₑₚᵣₑᵢ = v₀cos(θ), we can find:
v₀ₚₑₚᵣₑᵢ = 30 m/s * cos(50 degrees) = 19.27 m/s
Substituting these values into the equation for x:
x = (19.27 m/s) * (4.73 s) = 91.17 m
Therefore, the wall is located at a horizontal distance of 91.17 meters from the launch point. Since there are two possible solutions, one solution is at 91.17 meters to the right of the launch point, and the other solution is 91.17 meters to the left of the launch point.