Question

You and a friend stand on a snow-covered roof. you both throw snowballs from an elevation of 15 m with the same initial speed of 13 m/s, but in different directions. you throw your snowball downward, at 40° below the horizontal; your friend throws her snowball upward, at 40° above the horizontal. what is the speed of each ball when it is 5.0 m above the ground? (neglect air resistance.)

Answer 1

The speed of the snowball thrown downward when it is 5.0 m above the ground is 11.82 m/s, and the speed of the snowball thrown upward when it is 5.0 m above the ground is 13.44 m/s.

Step-by-step explanation:

To find the speed of each snowball when it is 5.0 m above the ground, we can use the equations of motion for projectiles. For the snowball thrown downward, we can use the equation:

vf = sqrt(vi^2 + 2gh)

where vf is the final velocity, vi is the initial velocity, g is the acceleration due to gravity, and h is the vertical distance. For the snowball thrown upward, we can use the same equation with a negative value for h. Substituting the given values:

For the snowball thrown downward: vf = sqrt((13 m/s)^2 + 2*(-9.8 m/s^2)×(15 m - 5 m)) = 11.82 m/s

For the snowball thrown upward: vf = sqrt((13 m/s)^2 + 2×(-9.8 m/s^2)×(-15 m + 5 m)) = 13.44 m/s

Answer 2

Refer to the diagram shown below.

g = 9.8 m/s², and air resistance is ignored.

Let us consider the two trajectories.

Case A; The ball is launched 40° below the horizontal
u = (13 m/s)*cos 40° = 9.9586 m/s, the horizontal velocity
v = -(13 m/s)*sin 40° = -8.3562 m/s

When the ball is at 5 m above ground, it would have traveled 10 m.
The vertical downward velocity is
V² = (-8.3562 m/s²)² + 2*(-9.8 m/s²)*(-10 m) = 265.826
V = +/-16.304 m/s => V = -16.304 m/s (downward)
The horizontal velocity remains unchanged.
The speed of the ball is
√[(-16.304)² + 9.9586²] = 19.105 m/s

Case B: The ball is launched 40° above the horizontal.
u = 9.9586 m/s, as in case A
v = 8.3562 m/s

When the ball is 5 m above ground, the vertical velocity, V, is given by
V² = (8.3562 m/s)² + 2*(-9.8 m/s²)*(-10 m) = 265.826
V = +/- 16.304 m/s
The speed of the ball is
√[(-16.304)² + 9.9586²] = 19.105 m/s

Surprisingly, both speeds are the same.

Answer:
The speed of the ball when it is 5 m above ground is 19.1 m/s (both cases).