Question

in olympic platform diving, the athletes dive from a platform that is 32.8 feet above the water. a diver leaves the platform with an initial vertical velocity of 4 feet per second.

for how many seconds is the diver in the air before reaching the water?

1.56 seconds
2.34 seconds
4.37 seconds
3.15 seconds

Explanation:

Step1: Recall the vertical motion formula

The height \( h \) of an object in vertical motion is given by the formula \( h = -16t^2 + v_0t + h_0 \), where \( h \) is the height at time \( t \), \( v_0 \) is the initial vertical velocity, \( h_0 \) is the initial height, and \( t \) is the time in seconds. The acceleration due to gravity is approximately \( 32 \) feet per second squared, so the coefficient of \( t^2 \) is \( -\frac{1}{2} \times 32=-16 \) (negative because it's acting downward).

Here, \( h_0 = 32.8 \) feet (initial height above water), \( v_0 = 4 \) feet per second (initial upward velocity, positive because it's upward), and we want to find \( t \) when \( h = 0 \) (when the diver reaches the water). So we set up the equation:

\( 0=-16t^2 + 4t + 32.8 \)

Step2: Rearrange the quadratic equation

Rewrite the equation in standard quadratic form \( ax^2+bx + c = 0 \). Here, \( a=-16 \), \( b = 4 \), \( c = 32.8 \). The quadratic formula is \( t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \).

First, calculate the discriminant \( D=b^2-4ac \):

\( D=(4)^2-4\times(-16)\times(32.8) \)

\( D = 16 + 4\times16\times32.8 \)

\( D=16+2099.2 \)

\( D = 2115.2 \)

Step3: Apply the quadratic formula

Now, substitute \( a=-16 \), \( b = 4 \), and \( D = 2115.2 \) into the quadratic formula:

\( t=\frac{-4\pm\sqrt{2115.2}}{2\times(-16)} \)

We have two solutions for \( t \) from the \( \pm \) sign. We discard the negative solution because time cannot be negative.

First, calculate \( \sqrt{2115.2}\approx45.99 \)

Then, for the positive root:

\( t=\frac{-4 - 45.99}{-32} \) (we take the minus sign in the numerator because \( -4+\sqrt{2115.2} \) would be positive, but let's check:

Wait, actually, the quadratic formula is \( t=\frac{-b\pm\sqrt{D}}{2a} \), so with \( a=-16 \), \( b = 4 \):

\( t=\frac{-4\pm\sqrt{2115.2}}{-32}=\frac{4\mp\sqrt{2115.2}}{32} \)

We need to take the solution where the numerator is positive (since time is positive). So we take the minus sign in the numerator:

\( t=\frac{4-\sqrt{2115.2}}{32} \) would be negative (since \( \sqrt{2115.2}\approx45.99>4 \)), so we take the plus sign in the numerator:

\( t=\frac{4 + 45.99}{32}=\frac{49.99}{32}\approx1.56 \)? Wait, no, that can't be right. Wait, maybe I made a mistake in the sign of the initial velocity. Wait, if the diver leaves the platform with an initial vertical velocity of 4 feet per second, is it upward or downward? In platform diving, usually, the initial velocity when leaving the platform is upward (like a jump), so \( v_0 = 4 \) (upward, positive). But when the diver is moving towards the water, the height decreases. Wait, maybe I messed up the sign of \( h \). Wait, when the diver reaches the water, \( h = 0 \), and the initial height \( h_0 = 32.8 \), so the equation is \( 0=-16t^2+4t + 32.8 \), which is correct.

Wait, let's recalculate the discriminant:

\( b^2-4ac=(4)^2-4\times(-16)\times(32.8)=16 + 2099.2 = 2115.2 \), correct.

\( \sqrt{2115.2}\approx45.99 \)

Then, \( t=\frac{-4\pm45.99}{-32} \)

For the two solutions:

1. \( t=\frac{-4 + 45.99}{-32}=\frac{41.99}{-32}\approx - 1.31 \) (discard, negative time)
2. \( t=\frac{-4 - 45.99}{-32}=\frac{-49.99}{-32}\approx1.56 \)

Wait, but this contradicts the options? Wait, maybe the initial velocity is downward? If the initial velocity is downward, then \( v_0=-4 \) (since upward is positive, downward is negative). Let's try that.

So the equation becomes \( 0=-16t^2-4t + 32.8 \)

Then, \( a=-16 \), \( b=-4 \), \( c = 32.8 \)

Discriminant \( D=(-4)^2-4\times(-16)\times(32.8)=16 + 2099.2 = 2115.2 \) (same discriminant)

Quadratic formula: \( t=\frac{4\pm\sqrt{2115.2}}{2\times(-16)} \)

Wait, no, \( t=\frac{-b\pm\sqrt{D}}{2a}=\frac{4\pm\sqrt{2115.2}}{-32} \)

Now, take the positive solution:

\( t=\frac{4 - 45.99}{-32}=\frac{-41.99}{-32}\approx1.31 \) (still not matching). Wait, maybe the formula is different? Wait, in feet, the formula for vertical motion is \( h(t)=-16t^2 + v_0t + h_0 \), where \( v_0 \) is initial velocity (upward positive). Let's check the options. The options include 1.56, 2.34, 4.37, 3.15. Wait, maybe I made a mistake in the initial height or velocity. Wait, the platform is 32.8 feet above water, initial vertical velocity 4 ft/s (upward). Let's re - calculate the quadratic equation:

\( -16t^2+4t + 32.8 = 0 \)

Multiply both sides by - 1: \( 16t^2-4t - 32.8 = 0 \)

Now, \( a = 16 \), \( b=-4 \), \( c=-32.8 \)

Discriminant \( D=(-4)^2-4\times16\times(-32.8)=16 + 2099.2 = 2115.2 \)

Quadratic formula: \( t=\frac{4\pm\sqrt{2115.2}}{32} \)

Take the positive root:

\( t=\frac{4 + 45.99}{32}=\frac{49.99}{32}\approx1.56 \)

Ah, so that's the positive root. So the time is approximately 1.56 seconds.

Answer:

1.56 seconds (the option: 1.56 seconds)