a projectile is launched at a velocity of 125m/s at an angle of 20 degrees. A building is 200m away. At what height above the ground will the projectile strike the building?
How long to hit the building?
Horizontal problem:
u = 125 cos 20
t = 200/u
the projectile is therefore in the air for t seconds, use that in the vertical problem
Vertical problem:
Vi = 125 sin 20
h = 0 + Vi t - 4.9 t^2
i don't understand the h= 0+Vit-4.9t^2 part where did the 4.9 come from ??
To find the height above the ground where the projectile will strike the building, we need to break it down into horizontal and vertical components.
Step 1: Horizontal Component
Since the projectile is launched at an angle of 20 degrees, we can find the horizontal component of its velocity using trigonometry. The horizontal velocity can be calculated as:
Horizontal velocity = Initial velocity × cosine(angle)
Horizontal velocity = 125 m/s × cos(20°)
Horizontal velocity = 125 m/s × 0.9397
Horizontal velocity ≈ 117.46 m/s
Step 2: Time of Flight
Now, we need to find the time it takes for the projectile to travel the horizontal distance to the building. We can use the horizontal distance and horizontal velocity to calculate it. The time of flight can be calculated as:
Time of flight = Horizontal distance / Horizontal velocity
Time of flight = 200 m / 117.46 m/s
Time of flight ≈ 1.703 s
Step 3: Vertical Component
Next, we need to find the vertical component of the projectile's velocity. We can calculate it using trigonometry. The vertical velocity can be calculated as:
Vertical velocity = Initial velocity × sine(angle)
Vertical velocity = 125 m/s × sin(20°)
Vertical velocity = 125 m/s × 0.3420
Vertical velocity ≈ 42.75 m/s
Step 4: Maximum Height
Since there is no vertical acceleration (ignoring air resistance and considering only the effect of gravity), the time of flight can be divided into two equal halves. Therefore, the projectile will reach its maximum height halfway through the time of flight. The maximum height can be calculated as:
Maximum height = (Vertical velocity²) / (2 × g)
where g is the acceleration due to gravity (approximately 9.8 m/s²)
Maximum height = (42.75 m/s)² / (2 × 9.8 m/s²)
Maximum height ≈ 92.21 m
Step 5: Height Above the Ground
Finally, we can find the height above the ground where the projectile will strike the building.
Height above the ground = Maximum height - Building height
Since the building height is not given, assuming the building starts from the ground level, the height above the ground where the projectile will strike the building is approximately 92.21 m.
To find the height above the ground where the projectile will strike the building, we can use the principles of projectile motion. We can break down the problem into horizontal and vertical components.
Let's consider the horizontal component first. The horizontal distance the projectile travels is 200 meters.
We can use the following formula to find the time of flight (t) for the projectile in the horizontal direction:
t = d / v
where
t = time of flight
d = horizontal distance
v = initial velocity
Substituting the values, we have:
t = 200 / 125
t ≈ 1.6 seconds
Now, let's consider the vertical component. We know the initial velocity (125 m/s) and the launch angle (20 degrees). We can decompose the initial velocity into its vertical and horizontal components.
The vertical component of the initial velocity (v_y) can be calculated as:
v_y = v * sin(θ)
where
v_y = vertical component of velocity
v = initial velocity
θ = launch angle
Substituting the values, we have:
v_y = 125 * sin(20)
v_y ≈ 42.86 m/s
Next, we need to calculate the time it takes for the projectile to reach its highest point, using the vertical component of velocity (v_y) and acceleration due to gravity (g).
The time to reach the highest point can be calculated as:
t_h = v_y / g
where
t_h = time taken to reach the highest point
v_y = vertical component of velocity
g = acceleration due to gravity (approximately 9.8 m/s²)
Substituting the values, we have:
t_h = 42.86 / 9.8
t_h ≈ 4.38 seconds
Since the time to reach the highest point is half of the total time, the total time of flight (t_total) can be calculated as:
t_total = 2 * t_h
t_total ≈ 2 * 4.38
t_total ≈ 8.76 seconds
Finally, we can calculate the height of the projectile above the ground using the vertical component of velocity (v_y) and the total time of flight (t_total).
The height (h) can be calculated as:
h = v_y * t_total - (1/2) * g * t_total²
Substituting the values, we have:
h = 42.86 * 8.76 - (1/2) * 9.8 * (8.76)²
h ≈ 374.7 meters
Therefore, the projectile will strike the building at a height of approximately 374.7 meters above the ground.