Suppose a ball is tossed straight up into the air from height 5 feet. What should be the initial velocity to have the ball reach its maximum height after 1 second? h(t) = -16.1t + Vt + H

To find the initial velocity \( V \) that will allow the ball to reach its maximum height after 1 second, we can use the given height function:

\[
h(t) = -16.1t^2 + Vt + H
\]

Where:
- \( h(t) \) is the height of the ball at time \( t \),
- \( t \) is the time in seconds,
- \( V \) is the initial velocity (which we need to find),
- \( H \) is the initial height, which is 5 feet.

The ball reaches its maximum height when the velocity becomes zero. The velocity is the derivative of the height function with respect to time, so:

\[
v(t) = \frac{d}{dt}[-16.1t^2 + Vt + H] = -32.2t + V
\]

At the maximum height, \( v(t) = 0 \). We're told that this happens after 1 second, so substitute \( t = 1 \) into the velocity equation:

\[
0 = -32.2(1) + V
\]

Solving for \( V \):

\[
V = 32.2
\]

So, the initial velocity required is \( 32.2 \, \text{feet/second} \).

Would you like more details on the process, or have any questions?

Here are 5 related questions you might find interesting:
1. How would the time to reach maximum height change if the initial velocity increased?
2. What would the height of the ball be at its maximum point?
3. How long will it take for the ball to hit the ground after it is thrown?
4. What would happen if the ball is thrown with a downward initial velocity?
5. How would air resistance affect the motion of the ball?

**Tip**: The maximum height in projectile motion occurs when the velocity momentarily becomes zero before the object starts descending.

Learn how to calculate the initial velocity required for a ball to reach its maximum height in 1 second, using projectile motion and differentiation. This detailed solution covers kinematic principles and is suitable for high school physics students in Grades 10-12.

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