A child throws a snowball with horizontal velocity of 18m/s directly toward a tree, from a distance of 9.0m and a height above the ground of 1.5m. if the time interval is 0.50s, at what height above the ground will the snow ballhit the tree
The horizontal distance to the tree is 9 m/18 m/s = 0.5 s, but they have already told you that.
During the itme of flight, the snowball falls vertically
H = (1/2) g t^2 = 4.9*0.5^2 = 1.23 m
The snowball started out 1.5 m above ground and ends up 1.5-1.23 = 0.27 m above ground
Well, isn't it snowball-tastic? Let's calculate the height above the ground where the snowball will hit the tree. Since the snowball is thrown horizontally, we can just focus on the vertical motion.
We have the initial vertical position (1.5m), the final vertical position (the height we want to find), the initial velocity in the vertical direction (0m/s), and the time interval (0.50s).
Using the equation:
final position = initial position + initial velocity * time + (1/2) * acceleration * time^2,
where the acceleration is due to gravity, which is approximately -9.8 m/s^2.
Substituting the values into the equation, we have:
final position = 1.5m + 0m/s * 0.50s + (1/2) * (-9.8 m/s^2) * (0.50s)^2.
After some mathematical juggling, we find that the snowball will hit the tree at a height of approximately 1.4 meters above the ground.
So, be careful where you stand if you're trying to dodge those snowballs. That tree might just give you some cold shoulder!
To find the height above the ground where the snowball hits the tree, we can use the kinematic equation:
y = yo + voy*t + (1/2)*a*t^2
Where:
y is the final vertical position (height above the ground where the snowball hits the tree)
yo is the initial vertical position (height above the ground from which the snowball is thrown)
voy is the initial vertical velocity (which is 0 since the snowball is thrown horizontally)
a is the vertical acceleration (which is equal to -9.8 m/s^2 for objects near the Earth's surface)
t is the time interval
Substituting the given values:
yo = 1.5m
voy = 0 m/s
a = -9.8 m/s^2
t = 0.50 s
y = 1.5m + 0m/s * 0.50s + (1/2) * -9.8m/s^2 * (0.50s)^2
y = 1.5m + 0m + (1/2) * -9.8m/s^2 * 0.25s^2
y ≈ 1.5m - 1.225m
y ≈ 0.275m
Therefore, the snowball will hit the tree at a height of approximately 0.275 meters above the ground.
To determine the height above the ground at which the snowball will hit the tree, we need to analyze the horizontal and vertical motion separately using the given information.
First, let's focus on the horizontal motion. We know that the horizontal velocity of the snowball is 18 m/s, and the distance between the child and the tree is 9.0 m. Since the horizontal velocity remains constant, we can use the formula:
Distance = Velocity × Time
In this case, the distance is 9.0 m, the velocity is 18 m/s, and the time is given as 0.50 s. Plugging these values into the formula:
9.0 m = 18 m/s × 0.50 s
Now, we solve for the unknown. Dividing both sides of the equation by 0.50 s:
9.0 m / 0.50 s = 18 m/s
The result is 18 m/s, which indicates that it takes 0.50 s for the snowball to travel 9.0 m horizontally.
Now, let's move on to the vertical motion. We know that the initial height above the ground is 1.5 m, and we want to find the final height at which the snowball hits the tree. Since we are dealing with free fall motion in the vertical direction, we can utilize the equation:
Final Height = Initial Height + Vertical Displacement
The vertical displacement can be determined using the equation for motion with constant acceleration:
Vertical Displacement = (Initial Velocity × Time) + (0.5 × Acceleration × Time^2)
Initially, the snowball is at rest vertically, so the initial velocity is 0 m/s. The acceleration due to gravity is approximately -9.8 m/s^2 (taking downward as negative). We are given a time interval of 0.50 s. Plugging these values into the formula:
Vertical Displacement = (0 m/s × 0.50 s) + (0.5 × -9.8 m/s^2 × (0.50 s)^2)
Simplifying further:
Vertical Displacement = 0 m + (-1.225 m) (rounded to three decimal places)
The vertical displacement is approximately -1.225 m. This indicates that the snowball falls 1.225 m downward during the given time interval.
To find the final height above the ground, we add the vertical displacement to the initial height:
Final Height = 1.5 m + (-1.225 m)
Final Height = 0.275 m (rounded to three decimal places)
Therefore, the snowball will hit the tree at a height of approximately 0.275 m above the ground.