A ball is hit at an angle of 36 degrees to the horizontal from 14 feet above the ground. If it hits the ground 26 feet away (in horizontal distance), what was the initial velocity? Assume there is no air resistance.
math - Henry, Friday, April 8, 2011 at 9:59pm
X = hor. = 26Ft.
r(hyp.) = X / cosA = 26 / cos36=32.1Ft.
Y = ver. = 32.1sin36 = 18.9Ft.
d(down) = Vo + 0.5gt^2 = 18.9,
0 + 4.9t^2 = 18.9.
t^2 = 3.86,
t = 1.96s.
Vf^2 = Vo^2 + 2gd = 0,
Vo^2 + 2*(-9.8)18.9 = 0,
Vo^2 - 370.44 = 0,
Vo^2 = 370.44,
Vo(ver.) = 19.2m/s.
Vo = 19.2 / sin36 = 32.7m/s.
Mr.Henry,
would you please explain this answer with a diagram and represtions.
hello arun~! i think i also answered this yesterday,, i'll just re-post my solution here. if there's something you didn't understand, feel free to ask. :)
first we observe that the motion of the ball is projectile and its trajectory (path followed by the ball) is shaped like a parabola. the horizontal distance travelled by the ball is thus given by:
R = (vo)^2 * sin (2*theta) / g
where
R = range or the horizontal distance
vo = initial velocity
theta = angle of release
g = acceleration due to gravity (9.8 m/s^2 = 32 ft/s^2)
substituting,
26 = (vo)^2 * sin(2*36) / 32
26 = (vo)^2 * 0.02972
26/0.02972 = (vo)^2
874.82 = (vo)^2
getting the squareroot of both sides,
vo = 29.58 ft/s
hope this helps~ :)
Certainly! Let's start by understanding the given information and breaking down the solution step by step.
We have a ball that is hit at an angle of 36 degrees to the horizontal from a height of 14 feet above the ground. The ball lands 26 feet away in the horizontal direction.
To find the initial velocity of the ball, we can use trigonometry and some kinematic equations.
First, let's define some variables:
- X: the horizontal distance covered by the ball (26 feet in this case)
- A: the launch angle (36 degrees)
- r (hyp.): the resultant distance traveled by the ball in a straight line (diagonal distance)
Now, let's calculate the horizontal distance (X) using the given value of 26 feet.
Next, we can calculate the horizontal component of the resultant distance (r) using the formula r (hyp.) = X / cos A. In this case, r = 26 / cos(36) = 32.1 feet.
Now, let's calculate the vertical distance (Y) covered by the ball. We can use the formula Y = r * sin A.
Substituting the values, we get Y = 32.1 * sin(36) = 18.9 feet.
Now, we can find the time it takes for the ball to reach the ground. We will use the equation d (down) = V₀ * t + 0.5 * g * t^2, where d (down) is the vertical distance covered, V₀ is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.
In this case, d (down) = 18.9 feet, V₀ = 0 (initial vertical velocity because the ball is hit horizontally), and g is approximately -9.8 m/s².
Using this equation, we can solve for time (t). Rearranging the equation, we get 0.5 * g * t^2 = d (down). Substituting the values, we have 0.5 * (-9.8) * t^2 = 18.9.
Simplifying, we find t^2 = 3.86, and then taking the square root, t ≈ 1.96 seconds.
Now, let's calculate the final vertical velocity (Vf) using the equation Vf^2 = V₀^2 + 2 * g * d (down).
Given that Vf is 0 (as the ball hits the ground) and d (down) is 18.9 feet, we can substitute these values into the equation. Vo is our unknown, so we'll solve for that.
We get 0 = Vo^2 + 2 * (-9.8) * 18.9. Simplifying, we find Vo^2 - 370.44 = 0, and then Vo^2 = 370.44.
Taking the square root, we find Vo (vertical component) ≈ 19.2 m/s.
Finally, we calculate the initial velocity (Vo) by dividing Vo (vertical component) by sin A.
So, Vo = 19.2 / sin(36) ≈ 32.7 m/s.
In summary, the initial velocity of the ball is approximately 32.7 m/s.
Apologies for not being able to provide a diagram in this text-based format. However, you can visualize the scenario by imagining a right-angled triangle with the angle of 36 degrees and the horizontal distance of 26 feet. The vertical distance covered by the ball is 18.9 feet, forming the triangle's opposite side. The diagonal distance covered by the ball (resultant distance) is 32.1 feet, forming the triangle's hypotenuse.