You throw a snowball at an angle of 34.0° below the horizontal from the roof of a building that is 74.0 m above flat ground. The snowball hits the ground a distance of 44.0 m from the base of the building. Assume free fall with up as positive, and use g = 9.80 m/s2.
how long the projectile remain in the air?
what is the initial speed of the snowball?
To find the time the projectile remains in the air, we can use the formula:
t = (2 * v0 * sin(theta)) / g
where:
v0 is the initial speed of the snowball
theta is the launch angle
g is the acceleration due to gravity
Given:
theta = 34.0°
g = 9.80 m/s^2
We need to find the initial speed (v0) of the snowball, and use that value to calculate the time (t).
First, let's find the horizontal and vertical components of the initial velocity.
Horizontal component:
v0x = v0 * cos(theta)
Vertical component:
v0y = v0 * sin(theta)
We know that the horizontal distance traveled by the snowball is 44.0 m and the vertical distance it traveled is 74.0 m. Let's use these values to solve for v0.
For the horizontal motion:
x = v0x * t
44.0 m = v0 * cos(theta) * t ---(i)
For the vertical motion:
y = v0y * t - (1/2) * g * t^2
74.0 m = v0 * sin(theta) * t - (1/2) * 9.80 m/s^2 * t^2 ---(ii)
From equation (i):
v0 * cos(theta) * t = 44.0 m
t = 44.0 m / (v0 * cos(theta))
Substituting the value of t in equation (ii):
74.0 m = v0 * sin(theta) * (44.0 m / (v0 * cos(theta))) - (1/2) * 9.80 m/s^2 * [(44.0 m / (v0 * cos(theta)))^2]
Now, we can solve this equation for v0 using algebraic methods.
To find the time the projectile remains in the air, we can use the vertical motion equation:
y = y0 + v0y * t - (1/2) * g * t^2
Where:
y is the vertical distance (in this case, the height of the building) = 74.0 m
y0 is the initial vertical position = 0 (ground level)
v0y is the vertical component of the initial velocity, which we need to solve for
g is the acceleration due to gravity = 9.80 m/s^2
t is the time in the air, which we want to find
Using the equation above, we can rearrange it to solve for t:
0 = 74.0 + v0y * t - (1/2) * 9.80 * t^2
Simplifying further, we get:
(1/2) * 9.80 * t^2 - v0y * t - 74.0 = 0
This is a quadratic equation in terms of t. We can solve it using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = (1/2) * 9.80 = 4.90, b = -v0y, and c = -74.0.
Now, to find the initial speed of the snowball, we can use the horizontal motion equation:
x = x0 + v0x * t
Where:
x is the horizontal distance = 44.0 m
x0 is the initial horizontal position = 0 (base of the building)
v0x is the horizontal component of the initial velocity, which we need to solve for
t is the time in the air, which we found earlier
Using this equation, we can rearrange it to solve for v0x:
v0x = x / t
So, now we need to calculate t using the quadratic equation we obtained earlier and then substitute it in to find v0x. Let's solve for t first:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
t = (-(-v0y) ± sqrt((-v0y)^2 - 4 * 4.90 * (-74.0))) / (2 * 4.90)
Dx = Vo^2*sin(2A)/g = 44. m.
Vo^2*sin(68)/9.8 = 44.
Vo^2*0.0946 = 44.
Vo^2 = 465.1.
Vo = 21.6 m/s. at 34o below the hor.
Xo = Vo*Cos A = 21.6*Cos34 = 17.9 m/s.
Dx = 17.9*T = 44.
T = 2.46 s. = Time in air.