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Sarah launches a snowball at

a rate of 16 feet per second
from a hill that is 32 feet off
the ground.
When does the
snowball hit the
ground?
What is the maximum
height the snowball
reaches?

User Verpous
by
8.6k points

1 Answer

4 votes

Explanation:

We can use the kinematic equations of motion to solve this problem. Let's assume the initial velocity of the snowball is 16 feet per second and its initial height is 32 feet. Also, we know that the acceleration due to gravity is -32.2 feet per second squared (assuming downward direction as negative).

To find out when the snowball hits the ground, we can use the equation:

h = 32 + 16t - 16t^2

where h is the height of the snowball at time t. We want to find the value of t when h = 0 (since the snowball hits the ground at that point). Therefore, we can rewrite the equation as:

16t^2 - 16t - 32 = 0

Dividing both sides by 16, we get:

t^2 - t - 2 = 0

Solving for t using the quadratic formula, we get:

t = (1 ± √(1 + 8))/2

t = 2 seconds or -1 second

Since time cannot be negative, the snowball hits the ground after 2 seconds.

To find the maximum height the snowball reaches, we can use the fact that the maximum height occurs at the vertex of the parabolic trajectory. The x-coordinate of the vertex is given by:

t = -b/2a

where a and b are the coefficients of the quadratic equation. In this case, a = 16 and b = -16, so:

t = -(-16)/(2*16) = 0.5 seconds

To find the corresponding height, we can substitute t = 0.5 seconds into the equation for h:

h = 32 + 16(0.5) - 16(0.5)^2

h = 36 feet

Therefore, the maximum height the snowball reaches is 36 feet.

User Karansardana
by
8.9k points

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