Snowballs are thrown with a speed of 11 m/s from a roof 6.0 m above the ground. Snowball A is thrown straight downward; snowball B is thrown in a direction 20^\circ above the horizontal.
1-Find the direction of motion of the two snowballs just before they land.
Well, let me tell you, these snowballs have quite the personalities! Snowball A, being thrown straight downward, is a real go-getter. It's all about that vertical motion, so its direction of motion just before landing will be straight down. It's a no-nonsense kind of snowball.
Now, Snowball B, on the other hand, is a bit of a daredevil. Thrown at an angle of 20 degrees above the horizontal, it's all about that curve. So, just before it lands, it will have a downward component due to gravity, but it will still have a bit of that horizontal motion. I like to think of it as Snowball B doing a little acrobatic dance move in the air before it lands.
So, to sum it up, Snowball A goes straight down, like a boss, while Snowball B does a fancy horizontal/vertical dance move. It's quite the spectacle, I must say!
To find the direction of motion of the two snowballs just before they land, we need to analyze their respective velocities.
For snowball A, which is thrown straight downward, the vertical component of its initial velocity will be negative, indicating a downward motion. The horizontal component will be zero since it is not thrown in a horizontal direction.
For snowball B, which is thrown in a direction 20° above the horizontal, we can break down its initial velocity into horizontal and vertical components. The vertical component will be positive since it is thrown above the horizontal, and the horizontal component will also be positive.
Since both snowballs will fall vertically, their directions of motion just before landing will be the same, which is downward.
To find the direction of motion of the two snowballs just before they land, we need to analyze their respective horizontal and vertical components of velocity.
For snowball A, which is thrown straight downward, the initial velocity is purely vertical since there is no horizontal motion. The magnitude of the vertical velocity is the same as the magnitude of the initial velocity, which is 11 m/s, but it is directed downward.
For snowball B, which is thrown 20 degrees above the horizontal, we need to split the initial velocity into horizontal and vertical components. The horizontal component can be found by multiplying the initial velocity by the cosine of the angle, while the vertical component can be found by multiplying the initial velocity by the sine of the angle.
Horizontal component of velocity (Vx) = 11 m/s * cos(20°)
Vertical component of velocity (Vy) = 11 m/s * sin(20°)
Now that we have the horizontal and vertical components of velocity for both snowballs, we can determine their directions of motion just before they land.
For snowball A, since it is only moving vertically downward, the direction of motion just before it lands is downward.
For snowball B, the direction of motion just before it lands can be determined by considering the sign of both the horizontal and vertical components of velocity. If both components are negative, then the direction of motion will be downward and to the left. If the horizontal component is positive and the vertical component is negative, then the direction of motion will be downward and to the right.
for ball A- it continues straight down
for ball B:
v= 11 sin 20 - g t
h = 6 + 11 sin 20 t - 4.9 t^2
when h = 0
0 = 6 + 3.76 t - 4.9 t^2
t^2 - .768 t - 1.22 = 0
t = [ .768 +/- sqrt (.59+4.88) ]/2
t = [ .768 +/- 2.33 ]/2
t = 1.55 seconds
v = 3.76 - 9.8 (1.55)
v = -11.43 vertical speed
u = 11 cos 20 = 10.33
so speed at h = 0 = sqrt(11.43^2+10.33^2)
and angle down from horizontal = tan^-1 (11.43/10.33)