1) 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.
i have posted the same question earlier and i got some steps to the answer .i am very thankful to you .Butwith that I am not able to reach the correct numericl answer which is 22.415 ft./sec
To find the initial velocity of the ball, we can break down the problem into horizontal and vertical components.
Let's start by analyzing the vertical motion of the ball. The initial vertical position is 14 feet above the ground, and the final vertical position is on the ground. The acceleration due to gravity, which acts vertically downward, is 32.2 ft/s^2 (approximately).
We can use the following kinematic equation to relate the initial vertical velocity, time of flight, and vertical displacement:
h = (v0y * t) - (0.5 * g * t^2),
where:
h is the vertical displacement (14 feet),
v0y is the initial vertical velocity,
t is the time of flight, and
g is the acceleration due to gravity (32.2 ft/s^2).
Rearranging the equation, we get:
v0y = (h + 0.5 * g * t^2) / t.
Now, let's move on to the horizontal motion of the ball. We are given that the horizontal distance traveled is 26 feet, and the angle of projection is 36 degrees with respect to the horizontal.
Using the equation for horizontal displacement, we can relate the initial horizontal velocity, time of flight, and horizontal displacement:
d = v0x * t,
where:
d is the horizontal displacement (26 feet), and
v0x is the initial horizontal velocity.
Since the horizontal velocity remains constant throughout the motion (no horizontal acceleration), we can find v0x as follows:
v0x = d / t.
To find the time of flight (t), we can use the following equation derived from the vertical motion:
t = sqrt((2 * h) / g).
Now, substituting the above values into the equations, we can find the initial vertical and horizontal velocities separately.
1. Calculate the time of flight (t):
t = sqrt((2 * 14) / 32.2) = 0.847 seconds (approximately).
2. Calculate the initial vertical velocity (v0y):
v0y = (14 + 0.5 * 32.2 * (0.847^2)) / 0.847 = 17.825 ft/s (approximately).
3. Calculate the initial horizontal velocity (v0x):
v0x = 26 / 0.847 = 30.679 ft/s (approximately).
To find the magnitude of the initial velocity, we can use the Pythagorean theorem:
v0 = sqrt(v0x^2 + v0y^2) = sqrt((30.679^2) + (17.825^2)) = 34.365 ft/s (approximately).
Hence, the initial velocity of the ball is approximately 34.365 ft/s.