A child throws a snowball with a horizontal velocity of 18 m/s directly toward a tree, from a distance of 9.0 m and a height above the ground of 1.5 m.
a) After what interval does the snowball hit the tree?
b) At what height above the ground will the snowball hit the tree?
c) Determine the snowball's velocity as it strikes the tree.
(a) time = distance/speed. The horizontal speed is constant.
(b) h(t) = 1.5 - 4.9t^2
(c) v^2 = 18^2 + (-9.8t)^2
To answer these questions, we can use the equations of motion to analyze the horizontal and vertical components of the snowball's motion separately.
a) To find the time interval, we need to determine the time it takes for the snowball to reach the tree. Since the initial horizontal velocity is 18 m/s and the horizontal displacement is 9.0 m, we can use the equation:
distance = velocity × time
9.0 m = 18 m/s × time
Solving for time:
time = 9.0 m / 18 m/s
time = 0.5 s
Therefore, the snowball takes 0.5 seconds to hit the tree.
b) To find the height above the ground where the snowball hits the tree, we need to consider the vertical motion of the snowball. We can use the equation of motion:
height = initial height + (initial vertical velocity × time) + (0.5 × acceleration × time²)
In this case, the initial height is 1.5 m, the initial vertical velocity is 0 m/s (since the snowball's motion is purely horizontal), and the acceleration is due to gravity, which is approximately -9.8 m/s² (assuming the positive upward direction).
Using the equation, we have:
height = 1.5 m + (0 m/s × 0.5 s) + (0.5 × (-9.8 m/s²) × (0.5 s)²
Simplifying:
height = 1.5 m + 0 + (-1.225 m)
height = 0.275 m
Therefore, the snowball hits the tree at a height above the ground of 0.275 m.
c) Since there is no vertical velocity at the time of impact, the snowball's velocity as it strikes the tree is purely horizontal. Therefore, the velocity is 18 m/s.