A person at the roof of a high building has projected a ball at velocity 50m/s given that the acceleration due to gravity = 10 the displacement and velocity when : The ball has been protected upwards with angle 30 to the horizontal. And when the ball been projected horizontally a zero degree
To find the displacement and velocity of the ball in both cases, we will use the equations of projectile motion.
1. Projectile motion with an angle of 30 degrees to the horizontal:
When the ball is projected upwards at an angle of 30 degrees, we need to find the vertical and horizontal components of the initial velocity.
Vertical Component:
The vertical component can be calculated using the formula:
Vy = V * sinθ
where V is the initial velocity (50 m/s) and θ is the launch angle (30 degrees).
Vy = 50 * sin(30) = 25 m/s
Horizontal Component:
The horizontal component can be calculated using the formula:
Vx = V * cosθ
where V is the initial velocity (50 m/s) and θ is the launch angle (30 degrees).
Vx = 50 * cos(30) = 43.3 m/s
Now, let's find the displacement and velocity at the topmost point of the ball's trajectory.
Time taken to reach maximum height:
The time taken to reach the maximum height can be calculated using the formula:
t = Vy / g
where Vy is the vertical component of the velocity (25 m/s) and g is the acceleration due to gravity (10 m/s^2).
t = 25 / 10 = 2.5 s
Maximum height reached:
The maximum height can be calculated using the formula:
h = (Vy^2) / (2g)
where Vy is the vertical component of the velocity (25 m/s) and g is the acceleration due to gravity (10 m/s^2).
h = (25^2) / (2 * 10) = 31.25 m
Displacement at the topmost point (vertical component):
The displacement can be calculated using the formula:
s = Vyt - (1/2)gt^2
where Vyt is the vertical component of the velocity at time t (25 m/s), g is the acceleration due to gravity (-10 m/s^2), and t is the time taken to reach maximum height (2.5 s).
s = 25 * 2.5 - (1/2) * 10 * (2.5^2) = 62.5 - 31.25 = 31.25 m
Displacement at the topmost point (horizontal component):
The horizontal displacement at the topmost point is equal to the initial horizontal velocity multiplied by the time taken to reach the maximum height.
s = Vx * t = 43.3 * 2.5 = 108.25 m
Therefore, the displacement of the ball at the topmost point is 31.25 m vertically upwards and 108.25 m horizontally.
2. Projectile motion with a horizontal projection (0 degrees angle):
When the ball is projected horizontally, the initial velocity in the vertical direction is zero (Vy = 0).
Vertical Component:
Vy = 0 m/s
Horizontal Component:
The horizontal component can be calculated using the formula:
Vx = V * cosθ
where V is the initial velocity (50 m/s) and θ is the launch angle (0 degrees).
Vx = 50 * cos(0) = 50 m/s
Time taken to reach the ground:
The time taken to reach the ground can be calculated using the formula:
t = 2 * (Vy / g), where Vy is zero in this case.
t = 2 * 0 / 10 = 0 s
Displacement in the vertical direction:
Since the initial velocity in the vertical direction is zero, the ball will fall vertically downwards.
Therefore, the displacement in the vertical direction can be calculated using the formula s = (1/2)gt^2.
s = (1/2) * 10 * (0^2) = 0 m
Displacement in the horizontal direction:
The horizontal displacement can be calculated using the formula:
s = Vx * t, where Vx is the horizontal component of velocity (50 m/s) and t is the time taken to reach the ground (0 s).
s = 50 * 0 = 0 m
Therefore, the displacement of the ball is 0 m vertically and horizontally when it is projected horizontally (at 0 degrees angle).
To find the displacement and velocity of the ball in each case, we can use the equations of motion.
1. When the ball is projected upwards at an angle of 30 degrees to the horizontal:
Initial velocity of the ball (u) = 50 m/s
Angle of projection (θ) = 30 degrees
Acceleration due to gravity (g) = 10 m/s^2
First, we need to resolve the initial velocity into its horizontal and vertical components:
Horizontal component of velocity (u_x) = u * cos(θ)
Vertical component of velocity (u_y) = u * sin(θ)
u_x = 50 * cos(30) = 43.30 m/s
u_y = 50 * sin(30) = 25.00 m/s
Next, we can calculate the time taken for the ball to reach maximum height:
Time to reach maximum height (t) = u_y / g
t = 25.00 / 10 = 2.50 seconds
Using this time, we can find the displacement of the ball at maximum height:
Vertical displacement at maximum height (h) = (u_y^2) / (2 * g)
h = (25.00^2) / (2 * 10) = 31.25 meters
Finally, we can find the total displacement and final velocity of the ball by using the equations of motion:
Total displacement (d) = 2 * h
d = 2 * 31.25 = 62.50 meters
Final velocity (v) = sqrt((u_x)^2 + (-u_y + g * t)^2)
v = sqrt((43.30)^2 + (-25.00 + 10 * 2.50)^2) = 46.09 m/s
Therefore, when the ball is projected upwards at an angle of 30 degrees, the displacement is 62.50 meters and the final velocity is 46.09 m/s.
2. When the ball is projected horizontally at zero degrees:
Initial velocity of the ball (u) = 50 m/s
Angle of projection (θ) = 0 degrees
Acceleration due to gravity (g) = 10 m/s^2
Since the ball is projected horizontally, the initial vertical component of velocity is zero:
Vertical component of velocity (u_y) = 0 m/s
The time taken for the ball to reach the ground can be found using the equation:
Time to reach the ground (t) = sqrt((2 * h) / g)
t = sqrt((2 * 0) / 10) = 0 seconds
Since the ball is projected horizontally, its vertical displacement is zero:
Vertical displacement (d_y) = 0 meters
The horizontal component of velocity remains the same as the initial velocity:
Horizontal component of velocity (u_x) = u * cos(θ)
u_x = 50 * cos(0) = 50 m/s
The total horizontal displacement can be calculated using the equation:
Total horizontal displacement (d_x) = u_x * t
d_x = 50 * 0 = 0 meters
Therefore, when the ball is projected horizontally at zero degrees, the displacement is 0 meters in both the vertical and horizontal directions, and the final velocity is 50 m/s (which remains constant throughout the motion).