Ryan throws a tennis ball straight up into the air. The ball reaches its maximum height at 2 seconds. The approximate height of the ball x seconds after being thrown is shown in the table.
Which equation models the motion of the ball?
To determine the equation that models the motion of the ball, we need to analyze the given information. Since the ball reaches its maximum height at 2 seconds, we can assume that the ball starts at a height of zero and returns to a height of zero at 4 seconds (as it goes up for 2 seconds and comes down for 2 seconds).
The table is not given, so we cannot analyze the information directly. However, based on the information provided, we can assume that the equation should be a quadratic equation since the motion of an object thrown vertically up and then falls back down is typically represented by an equation of that form.
In general, the equation for the height of the object at a given time is modelled by the quadratic equation:
h(t) = -16t^2 + vt + h0
where h(t) represents the height at time t, v is the initial velocity of the object (going upwards in this case), and h0 is the initial height (which is zero in this scenario).
Without the table, we cannot determine the exact values of v and h0, but we can conclude that the equation representing the motion of the ball should be of the form:
h(t) = -16t^2 + vt
where v is the initial velocity.
To determine the equation that models the motion of the ball, we need to analyze the given information and observe any patterns.
From the information provided, we know that the ball reaches its maximum height at 2 seconds. This indicates that the ball takes 2 seconds to reach its peak and start descending.
The table gives us the approximate height of the ball at various time intervals. Let's look for a pattern in the height values. It appears that the height decreases symmetrically after reaching the maximum value.
Based on this observation, we can conclude that the equation representing the motion of the ball will be a quadratic equation. A general quadratic equation can be expressed in the form:
h(t) = at^2 + bt + c
Here, h(t) represents the height of the ball at time t, and a, b, and c are constants.
To determine the values of the constants, we need to use the given information. Since the maximum height is reached at 2 seconds, we know that h(2) will be the maximum height of the ball.
From the table, we can see that the maximum height is 16 units. Plugging these values into the equation, we get:
16 = a(2)^2 + b(2) + c
16 = 4a + 2b + c
Now, let's look at the other values in the table. At t = 0, the height is 10 units. Plugging these values into the equation, we get:
10 = a(0)^2 + b(0) + c
10 = c
Now, we have determined the value of c as 10.
Substituting this into the equation, we get:
16 = 4a + 2b + 10
6 = 4a + 2b
We now have two equations with two unknowns:
4a + 2b = 6
a + b = 3
Solving this system of equations, we find that a = 1 and b = 2.
Therefore, the equation that models the motion of the ball is:
h(t) = t^2 + 2t + 10
let the initial velocity be k m/s
so
height = -4.9t^2 + kt
d(height)/dt = -9.8t + k
when t = 2, d(height)/dt = 0
2 = -9.8(2) + k
k = 19.6
so your equation was
height = -4.9t^2 + 19.6t
the maximum height is
= -4.9(4) + 19.6(2) m
= appr 19.6 m
I used the more traditional variable t for time instead of the suggested x, no big deal.
well, you know that
y = vt - 4.9t^2
max height is at t = v/9.8