A ball is thrown directly upward from the ground with an initial velocity of 4.8 ft./sec. Represent the height of the ball from the ground t seconds after it was thrown upward using the model h(t) .(1 point)
1. h(t) = -16t^2 + 4.8t
2. h(t) = -1/2 x 32t^2 + 4.8t + 4.8
3. h(t) = -1/2 x 9.8t^2 + 4.8t
4. h(t) = -1/2 x 4.8t^2 + 4.8t
The correct answer is 4. h(t) = -1/2 x 4.8t^2 + 4.8t.
The correct representation of the height of the ball from the ground t seconds after it was thrown upward is option 4: h(t) = -1/2 x 4.8t^2 + 4.8t.
To determine the correct model for the height of the ball from the ground at time t, we can use the following steps:
Step 1: Identify the relevant formula.
In this case, we are dealing with an object that is thrown directly upward, so we need to use the equation for vertical motion under gravity:
h(t) = -16t^2 + v0t + h0,
where h(t) represents the height of the ball at time t, v0 is the initial velocity, and h0 is the initial height.
Step 2: Plug in the given values.
Given that the ball is thrown upward from the ground with an initial velocity of 4.8 ft./sec, we have:
v0 = 4.8 ft/sec
h0 = 0 ft (since the ball is thrown from the ground)
Step 3: Substitute the values into the formula.
Plugging in the values for v0 and h0 into the formula, we get:
h(t) = -16t^2 + 4.8t + 0.
Simplifying, we have:
h(t) = -16t^2 + 4.8t.
Therefore, the correct model for the height of the ball from the ground at time t is:
h(t) = -16t^2 + 4.8t.
Answer: Option 1. h(t) = -16t^2 + 4.8t.