A cannon ball is fired straight up into the air at a speed of 80 ft/sec from a height of 2 ft. a)when will be the cannon ball be 98 feet in the air. b) when will the cannon ball hit the ground?
the equation for the ball's height at time t will be
h = 2 + 80t - 16t^2
So, it will be 98' up when t = 2
h=0 when t = 5.025 sec
(at t=5, it will be back to its original 2' height)
To solve this problem, we can use the equations of motion and kinematics. Let's break the problem down into two parts.
a) When will the cannon ball be 98 feet in the air?
First, we need to find the time it takes for the cannon ball to reach a height of 98 feet. We'll use the equation:
h(t) = -16t^2 + v₀t + h₀
where h(t) is the height, v₀ is the initial velocity, t is time, and h₀ is the initial height.
Plugging in the given values:
h(t) = -16t^2 + 80t + 2
To find when the ball is 98 feet in the air, we set h(t) equal to 98:
-16t^2 + 80t + 2 = 98
Simplifying the equation:
-16t^2 + 80t - 96 = 0
Dividing the equation by -16:
t^2 - 5t + 6 = 0
Factoring the quadratic equation:
(t - 2)(t - 3) = 0
Therefore, t can be either 2 or 3. However, since the cannon ball would have already been at 98 feet on its way down at t = 2, we can conclude that the cannon ball would be at a height of 98 feet in the air at t = 3 seconds.
b) When will the cannon ball hit the ground?
To find when the cannon ball hits the ground, we need to find the time at which the height becomes zero.
Setting h(t) = 0, we have:
-16t^2 + 80t + 2 = 0
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values:
t = (-80 ± √(80^2 - 4(-16)(2))) / (2(-16))
t = (-80 ± √(6400 + 128)) / -32
t = (-80 ± √(6528)) / -32
t ≈ (-80 ± 80.99) / -32
Simplifying:
t ≈ (-80 + 80.99) / -32 or t ≈ (-80 - 80.99) / -32
t ≈ 0.03 or t ≈ 5.03
Therefore, the cannonball will hit the ground at approximately t = 5.03 seconds.
In summary, the cannonball will be 98 feet in the air at t = 3 seconds and it will hit the ground at approximately t = 5.03 seconds.