A tennis ball is thrown horizontally at a speed of 10 m/s from the top of a building 78.4 meters high. How long does it take the ball to hit the street?

To find out how long it takes for the tennis ball to hit the street, we can use the equation of motion:

\[ h = ut + \frac{1}{2}gt^2 \]

where:
- h is the height of the building (78.4 meters),
- u is the initial horizontal velocity of the ball (10 m/s),
- g is the acceleration due to gravity (-9.8 m/s^2),
- t is the time it takes for the ball to hit the street (unknown).

Since the ball is thrown horizontally, its initial vertical velocity is 0 m/s. Therefore, we can ignore the first term in the equation.

Now let's rearrange the equation to solve for t:

\[ h = \frac{1}{2}gt^2 \]

\[ \frac{2h}{g} = t^2 \]

\[ t = \sqrt{\frac{2h}{g}} \]

Plugging in the values, we have:

\[ t = \sqrt{\frac{2 \times 78.4}{9.8}} \]

\[ t = \sqrt{\frac{156.8}{9.8}} \]

\[ t = \sqrt{16} \]

\[ t = 4 \]

Therefore, it takes the tennis ball 4 seconds to hit the street.

To determine how long it takes for the tennis ball to hit the street, we can break down the problem into two parts: the vertical motion and the horizontal motion.

First, let's focus on the vertical motion. The initial height of the ball is 78.4 meters, and we need to find the time it takes to reach the street, which would be when the height is 0.

Using the equation for vertical motion, which is h = ut + (1/2)gt^2, where:
h is the height (78.4 m),
u is the initial vertical velocity (0 m/s),
t is the time, and
g is the acceleration due to gravity (-9.8 m/s^2),

we can rearrange the equation to solve for time:

0 = 0 * t + (1/2)(-9.8)t^2

Since the initial vertical velocity is 0, the first term becomes 0. Simplifying the equation, we have:

0 = (-4.9)t^2

Now, we can solve for time:

(-4.9)t^2 = 0

t^2 = 0

t = 0

Since time cannot be zero in this case, this means that the equation has no real solution for time. Therefore, the vertical motion of the ball does not play a role in determining how long it takes to hit the street.

Now let's consider the horizontal motion. The ball is thrown horizontally at a speed of 10 m/s. This means that the horizontal velocity remains constant throughout the motion.

To find the time it takes for the ball to hit the street horizontally, we can use the equation d = vt, where:
d is the horizontal distance (unknown),
v is the horizontal velocity (10 m/s),
and t is the time (unknown).

Since we know the height of the building, we can use it as the horizontal distance d. Therefore, we can rewrite the equation as:

78.4 m = 10 m/s * t

To solve for time, we divide both sides of the equation by 10 m/s:

t = 78.4 m / 10 m/s

Calculating the result, we get:

t = 7.84 s

So, it takes approximately 7.84 seconds for the tennis ball to hit the street.

Hint:

The answer has nothing to do withe the 10 m/s horizontal velocity com[ponent.

Just calculate how long it takes for an object to fall 78.4 m

Y = (1/2) g t^2

(ignoring air resistance)

Solve for t.

Exactly as long as it takes to fall 78.4 meters since it had zero initial vertical velocity (only horizontal).

(1/2)(9.8) t^2 = 78.4