A ball is thrown at 41° below the horizontal from a rooftop of height 65 m. It lands 2 s later. Find: Find:
a) its initial speed
b) the point of impact with the ground;
m from the base of the building
c) the angle at which it hits the ground.
° below the horizontal
To solve this problem, we can use the equations of motion for projectile motion. Projectile motion is the motion of an object in a two-dimensional plane under the influence of gravity.
Let's start by finding the initial speed of the ball.
a) The initial speed of the ball can be found using the horizontal and vertical components of its motion. The horizontal component remains constant throughout the motion, while the vertical component experiences uniform acceleration due to gravity.
Given:
- Angle below the horizontal = 41°
- Height of the rooftop = 65 m
- Time of flight = 2 s
To find the initial speed, we need to find the horizontal component of the velocity.
Horizontal component of the velocity (Vx) can be given by:
Vx = V * cos(theta)
where V is the initial speed and theta is the angle below the horizontal.
Vertical component of the velocity (Vy) can be given by:
Vy = V * sin(theta)
The time of flight can be broken down into two parts: the time taken to reach the maximum height (t1) and the time taken to come down from the maximum height to the ground (t2). Since the ball takes the same amount of time to reach the maximum height and to come down, we can write:
t1 = t2 = t/2
The height of the ball at the maximum height will be equal to the height of the rooftop.
Using the equation for vertical displacement (H) in projectile motion:
H = Vy * t1 + (1/2) * g * t1^2
Since the initial velocity in the vertical direction is zero (ball thrown from the rooftop vertically downward), we can rewrite the equation as:
H = (1/2) * g * t1^2
Substituting the value of t1 from above equation, we get:
65 = (1/2) * g * (t/2)^2
Simplifying the equation, we get:
65 = (1/8) * g * t^2
Now, we can find t from the given time of flight:
2 = t
Substituting the value of t in the equation, we get:
65 = (1/8) * g * 2^2
65 = (1/8) * g * 4
g = (65 * 8)/4
The acceleration due to gravity (g) is approximately 9.8 m/s^2.
Using the equation for horizontal displacement (D) in projectile motion:
D = V * cos(theta) * t
Since the ball is thrown from the rooftop horizontally, the initial horizontal velocity is zero. Therefore, the equation simplifies to:
D = 0 * cos(theta) * t
D = 0
The point of impact with the ground is at a horizontal distance of 0 m from the base of the building.
c) The angle at which the ball hits the ground can be found using the equation for angle of projection (alpha):
tan(alpha) = Vy / Vx
In this case, Vy = V * sin(theta) and Vx = V * cos(theta).
Using the values of Vy and Vx, we get:
tan(alpha) = (V * sin(theta)) / (V * cos(theta))
tan(alpha) = sin(theta) / cos(theta)
From the given information, theta = 41°. We can find tan(theta) using a scientific calculator or trigonometric tables.
Now, you can use the formulas explained above to find the answers to your question.