Q Exactly 2.7s after a projectile is fired into the air from the ground ,it is observed to have a velocity v = (8.8i+4.5 j) m/s where the x-axis is horizontal and the y axis is positive upward .
a)Determine the horizontal range of the projectile
b)Determine the maximum height above the ground
c)Determine the speed of motion just above the projectile strikes the ground .
d)Determine the angle of the motion just before the projectile strikes the ground .
a) The horizontal range of the projectile can be determined by using the formula R = Vx * t, where Vx is the horizontal component of the velocity and t is the time. In this case, Vx = 8.8 m/s and t = 2.7 s. So, R = 8.8 m/s * 2.7 s = 23.76 m.
b) The maximum height above the ground can be determined by using the formula h = Vy * t + 0.5 * g * t^2, where Vy is the vertical component of the velocity, t is the time, and g is the acceleration due to gravity. In this case, Vy = 4.5 m/s, t = 2.7 s, and g = 9.8 m/s^2. So, h = 4.5 m/s * 2.7 s + 0.5 * 9.8 m/s^2 * (2.7 s)^2 = 16.37 m.
c) The speed of motion just above the projectile strikes the ground can be determined by using the Pythagorean theorem. The speed is the magnitude of the velocity vector, which is given by |v| = sqrt(Vx^2 + Vy^2). In this case, Vx = 8.8 m/s and Vy = 4.5 m/s. So, |v| = sqrt(8.8^2 + 4.5^2) = 9.93 m/s.
d) The angle of the motion just before the projectile strikes the ground can be determined by using the tangent function. The angle is given by tan(theta) = Vy / Vx. In this case, Vy = 4.5 m/s and Vx = 8.8 m/s. So, theta = tan^(-1)(4.5 / 8.8) = 27.43 degrees.
To solve this problem, we can use the equations of motion for projectile motion.
Step 1: Determine the horizontal range of the projectile
The horizontal range is the total horizontal distance covered by the projectile. It can be calculated using the equation:
Range = Horizontal velocity * Time of flight
Given that the horizontal velocity is given as 8.8 m/s, we can calculate the time of flight using the equation:
Time of flight = 2.7 s
Plugging in the values, we get:
Range = 8.8 m/s * 2.7 s
Range = 23.76 meters
Therefore, the horizontal range of the projectile is 23.76 meters.
Step 2: Determine the maximum height above the ground
To find the maximum height above the ground, we can use the equation:
Maximum height = (Vertical velocity^2) / (2 * acceleration due to gravity)
Given that the vertical velocity is given as 4.5 m/s, and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the maximum height:
Maximum height = (4.5 m/s)^2 / (2 * 9.8 m/s^2)
Maximum height = 0.102 meters
Therefore, the maximum height above the ground is 0.102 meters.
Step 3: Determine the speed of motion just before the projectile strikes the ground
The speed of motion just before the projectile strikes the ground can be calculated using the equation:
Speed = Square root of [(Horizontal velocity)^2 + (Vertical velocity)^2]
Given that the horizontal velocity is 8.8 m/s, and the vertical velocity is 4.5 m/s, we can calculate the speed:
Speed = Square root of [(8.8 m/s)^2 + (4.5 m/s)^2]
Speed = Square root of [77.44 + 20.25]
Speed = Square root of 97.69
Speed = 9.88 m/s
Therefore, the speed of motion just before the projectile strikes the ground is 9.88 m/s.
Step 4: Determine the angle of motion just before the projectile strikes the ground
To find the angle of motion just before the projectile strikes the ground, we can use the equation:
Angle = atan(Vertical velocity / Horizontal velocity)
Given that the vertical velocity is 4.5 m/s and the horizontal velocity is 8.8 m/s, we can calculate the angle:
Angle = atan(4.5 m/s / 8.8 m/s)
Angle = atan(0.5114)
Angle = 26.12°
Therefore, the angle of motion just before the projectile strikes the ground is 26.12°.
To find the answers to these questions, we need to apply the principles of motion in projectile motion. Let's break down the steps to solve each part of this problem:
a) Determine the horizontal range of the projectile:
The horizontal range is the total horizontal distance covered by the projectile. To find it, we need to calculate the time it takes for the projectile to reach the ground.
We know that the time is 2.7 seconds after the projectile is fired. The formula for horizontal distance is given by:
Range = horizontal velocity * time
Since the horizontal velocity is given as (8.8i + 4.5j) m/s, we can determine the horizontal component of velocity as 8.8 m/s.
Using the given time of 2.7 seconds, we can calculate the horizontal range as:
Range = 8.8 m/s * 2.7 s = 23.76 meters
Therefore, the horizontal range of the projectile is 23.76 meters.
b) Determine the maximum height above the ground:
The maximum height is reached when the vertical velocity of the projectile becomes zero. We need to find the time taken to reach maximum height.
The vertical velocity is given as 4.5 m/s. As the projectile goes up, the vertical velocity decreases due to gravity, until it reaches zero at maximum height. We can calculate the time taken to reach maximum height using the formula:
Time taken to reach maximum height = vertical velocity / acceleration due to gravity
Using the given vertical velocity of 4.5 m/s and taking the acceleration due to gravity as 9.8 m/s^2, we can calculate the time as:
Time = 4.5 m/s / 9.8 m/s^2 = 0.459 seconds
To find the maximum height, we can use the formula:
Height = vertical velocity * time - 0.5 * acceleration due to gravity * (time^2)
Height = 4.5 m/s * 0.459 s - 0.5 * 9.8 m/s^2 * (0.459 s)^2
Height = 2.0661 meters
Therefore, the maximum height above the ground is 2.0661 meters.
c) Determine the speed of motion just above the projectile strikes the ground:
At the moment just before the projectile strikes the ground, only the vertical component of velocity will be non-zero, as the horizontal component does not affect the speed at this point.
Therefore, to determine the speed just above the projectile strikes the ground, we only need to find the magnitude of the vertical velocity, which is already given as 4.5 m/s.
Therefore, the speed just above the projectile strikes the ground is 4.5 m/s.
d) Determine the angle of the motion just before the projectile strikes the ground:
To find the angle, we need to calculate the tangent of the angle based on the given velocity components.
The angle θ can be found using the formula:
θ = arctan(vertical velocity / horizontal velocity)
θ = arctan(4.5 m/s / 8.8 m/s)
θ ≈ 26.02°
Therefore, the angle of the motion just before the projectile strikes the ground is approximately 26.02°.
a) The horizontal component of any velocity remains constant as it goes through the air, so the horizontal component of the initial velocity would still be 8.8i. The distance can be calculated using the formula Xf=Vx*t:
Xf=8.8*2.7=23.76
b)Maximum height is given by the formula:
H_max=((Vo*sin(a)))^2/2g
The initial velocity can be found by finding the initial y component using the formula Vf=Vo+at
4.5=Vy-9.81*2.7=>Vy=30.987
Now Vo=sqrt(Vx^2+Vy^2)=sqrt(8.8^2+30.987^2)
Now you can find the angle between the horizontal and the firing velocity and plug it in the to formula given above.
c) and d) after the projectile as travelled the entire trajectory back to the ground, the speed and the angle the projectile has is equal to the initial velocity and angle