A golf ball with an initial angle of 35° lands exactly 218 m down the range on a level course.
(a) Neglecting air friction, what initial speed would achieve this result?
m/s
(b) Using the speed determined in item (a), find the maximum height reached by the ball.
m
Vertical equation
hf=hi+Vi*sin35*time - 4.9 time^2
0=t(Vi*sin35 -4.9t) so t is
t=Vi*sin35/4.9
Horizontal equation. Put that time t in to this equation, and solve for Vi.
218=vi*cos35*t
To solve this problem, we can break it down into two separate parts: the horizontal and vertical motion of the golf ball.
(a) To find the initial speed of the golf ball, we need to consider only the horizontal motion because there is no air friction.
We can use the equation for horizontal distance traveled:
d = v₀ * t
where
d = 218 m (horizontal distance traveled)
v₀ = initial velocity of the golf ball (to be determined)
t = time of flight
In projectile motion, the time of flight is given by:
t = 2 * v₀ * sin(θ) / g
where
θ = angle of projection (35°)
g = acceleration due to gravity (9.8 m/s²)
Combining the two equations above, we can find the initial velocity:
d = (2 * v₀ * sin(θ) / g) * v₀
218 = (2 * v₀ * sin(35°) / 9.8) * v₀
Simplifying and solving for v₀:
v₀² = 218 * 9.8 / (2 * sin(35°))
v₀ ≈ √(218 * 9.8 / (2 * sin(35°)))
Calculating in the next step:
(b) To find the maximum height reached by the ball, we need to consider only the vertical motion.
The maximum height can be calculated using the equation:
h_max = (v₀ * sin(θ))² / (2 * g)
Using the value of v₀ calculated in part (a):
h_max = (√(218 * 9.8 / (2 * sin(35°))) * sin(35°))² / (2 * 9.8)
Calculating in the next step:
To solve this problem, we'll use the principles of projectile motion. Projectile motion is a form of motion where an object is thrown near the Earth's surface, and the only forces acting on it are gravity and air resistance (which we'll neglect in this case).
(a) To determine the initial speed of the golf ball, we can use the kinematic equation for the horizontal range (x-axis):
Range = ((v^2) * sin(2θ)) / g
where:
- Range is the horizontal distance the ball travels (218 m),
- v is the initial speed of the ball,
- θ is the initial angle the ball is launched (35°), and
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
Rearranging the equation to solve for v, we have:
v = (√(Range * g)) / sin(2θ)
Plugging in the given values:
v = (√(218 * 9.8)) / sin(2 * 35°)
≈ (√(2136.4)) / sin(70°)
≈ √(2136.4) / 0.9397
≈ 46.14 / 0.9397
≈ 49.09 m/s
So, neglecting air friction, the initial speed required for the golf ball to land 218 m down the range is approximately 49.09 m/s.
(b) To calculate the maximum height reached by the ball, we'll use the vertical motion equation for projectile motion:
Height = (v^2 * (sin(θ))^2) / (2g)
where:
- Height is the maximum height,
- v is the initial speed of the ball (49.09 m/s),
- θ is the initial angle the ball is launched (35°), and
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the given values:
Height = (49.09^2 * (sin(35°))^2) / (2 * 9.8)
≈ (2402.1481 * (0.5736)^2) / 19.6
≈ (2402.1481 * 0.3297) / 19.6
≈ 792.8 / 19.6
≈ 40.41 m
Therefore, the maximum height reached by the golf ball is approximately 40.41 m.