a golf ball is observed to hit the ground 800 ft. away after a flight of 5 seconds. Assume that the ground is level and neglect friction of the air. Find the angle of elevation above the ground of the initial motion of the ball, find the initial velocity of the ball, find its maximum height above t he ground v = v + gt v = 160 ft/sec ?
or 800 = v(5) + 1/2(32)(25) = 240 ft/sec
D = Xo * t = 800 Ft.
Xo * 5 = 800
Xo = 160 Ft/s.
Tr + Tf = 5 s. Tr = Tf.
Tr+Tr = 5.
Tr = 2.5 s. = Rise time.
Y = Yo + g*Tr = 0.
Yo = -g*Tr = -(-32)*2.5 = 80 Ft/s.
1. Tan A = Yo/Xo = 80/160 = 0.50.
A = 26.6o.
2. Vo = Xo + Yoi = 160 + 80i = 179Ft/s[26.6o].
3. h = Yo*Tr + 0.5g*Tr^2.
g = (-)32 Ft/s^2.
Solve for h.
To find the angle of elevation and initial velocity of the golf ball, we can use the equations of motion for projectile motion.
Let's start by considering the horizontal motion of the golf ball. The horizontal distance traveled by the ball can be calculated using the equation:
horizontal distance (d) = initial velocity (v₀) * time (t)
In this case, the horizontal distance is given as 800 ft, and the time is 5 seconds. We can rearrange the equation to solve for the initial velocity:
v₀ = d / t
Plugging in the values, we get:
v₀ = 800 ft / 5 s = 160 ft/s
So the initial velocity of the golf ball is 160 ft/s.
Now, let's consider the vertical motion of the golf ball. We can use the equation for the vertical displacement to find the maximum height reached by the ball:
vertical displacement (h) = initial vertical velocity (v₀y) * time (t) - 1/2 * acceleration due to gravity (g) * time^2
Since the golf ball is hitting the ground 800 ft away after 5 seconds, we know that the vertical displacement is 0 (i.e., the ground is at the same height as the starting point). We also know that the acceleration due to gravity is approximately 32 ft/s^2. Rearranging the equation, we get:
0 = v₀y * 5 s - (1/2) * 32 ft/s^2 * (5 s)^2
Simplifying, we can solve for the initial vertical velocity, v₀y:
v₀y = (1/2) * 32 ft/s^2 * (5 s) / 5 s
v₀y = 80 ft/s
Now, let's find the angle of elevation of the initial motion of the ball. We can use the trigonometric relationship between the vertical and horizontal velocities:
tan(θ) = v₀y / v₀
θ = atan(v₀y / v₀)
Plugging in the values, we get:
θ = atan(80 ft/s / 160 ft/s)
θ ≈ atan(0.5)
θ ≈ 26.57 degrees
So, the angle of elevation above the ground of the initial motion of the ball is approximately 26.57 degrees.