A ball is thrown at an angle of 45 degrees to the ground. If the ball lands 90 m away, what is the initial speed of the ball?
To find the initial speed of the ball, we can use the horizontal and vertical components of the motion.
Given:
- The angle of projection, θ = 45 degrees.
- The horizontal distance covered, x = 90 m.
Step 1: Decompose the initial velocity into horizontal and vertical components.
Since the angle of projection is 45 degrees, the vertical and horizontal components of the initial velocity (Vi) are equal.
So, the vertical component (Vy) of Vi = horizontal component (Vx) of Vi.
Step 2: Calculate the horizontal component of the velocity.
Using the formula for horizontal distance covered:
x = Vx * t,
where Vx is the horizontal component of the velocity, and t is the time of flight.
We can rewrite this equation as:
Vx = x / t.
Step 3: Calculate the time of flight.
The time of flight for a projectile can be calculated using the formula:
t = 2 * (Vy / g),
where Vy is the vertical component of the velocity, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since Vy = Vx, we can rewrite the equation as:
t = 2 * (Vx / g).
Step 4: Substitute the values into the equation.
Using the equation Vx = x / t, we can substitute the values of x and t into the equation:
Vx = 90 / (2 * (Vx / g)).
Step 5: Simplify the equation.
By simplifying the equation, we get:
Vx^2 = 90 * g.
Step 6: Solve for Vx.
Taking the square root of both sides of the equation, we get:
Vx = sqrt(90 * g).
Since Vx = Vy, the initial speed of the ball can be calculated by multiplying either Vx or Vy by √2 (approximately 1.414).
Therefore, the initial speed of the ball is:
Vi = Vx * √2 = sqrt(90 * g) * √2.
Note: To get the numerical value of the initial speed, g needs to be provided in m/s^2.
To find the initial speed of the ball, we can use the kinematic equations of motion.
First, let's consider the motion in the horizontal direction. The horizontal motion of the ball is independent of the vertical motion, so we can ignore any effects of gravity in this direction.
We are given that the ball lands 90 m away, which is the horizontal displacement (x). The only other value we need is the time of flight (t), which is the total time taken by the ball to travel from the initial point to the landing point.
We can use the formula for horizontal displacement (x) to find the time of flight:
x = v0 * t * cos(θ)
Where:
v0 is the initial velocity (speed) of the ball
θ is the angle at which it is thrown (45 degrees in this case)
Since the angle of projection is 45 degrees, the value of cos(45) is 1/√2.
Therefore, we can rearrange the equation to solve for the time of flight (t):
t = x / (v0 * cos(θ))
Substituting the given values:
t = 90 / (v0 * 1/√2)
t = 90√2 / v0
Now, let's consider the vertical motion of the ball. We can use the formula for vertical displacement (y) to find the time of flight (t) again:
y = v0 * t * sin(θ) - (1/2) * g * t^2
Where:
y is the vertical displacement (which is zero in this case since the ball lands on the same level as it was thrown from)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Since the ball lands on the same level, we can rewrite the equation as:
0 = v0 * t * sin(θ) - (1/2) * g * t^2
Substituting the values, we get:
0 = v0 * (90√2 / v0) * sin(45) - (1/2) * 9.8 * (90√2 / v0)^2
Simplifying the equation will give us the value of v0, which is the initial speed of the ball.
Dx = Vo^2*sin(2A)/g = 90 m.
Vo^2*sin90 / 9.8 = 90.
Vo^2*0.102 = 90.
Vo^2 = 90 / 0.102 = 882.
Vo = 29.7 m/s.