Question

A stunt man jumps horizontally from a building to the roof of a garage that is 2 meters lower. How fast does he need to be to land on the roof of the said garage that is 3 meters away from the building?

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Eliseo

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To solve this problem, we can use the principles of kinematics and the equation of motion.

Let's assume the initial velocity of the stunt man isv_0 and the time taken to reach the roof of the garage is t . Since the stunt man is jumping horizontally, the vertical component of his velocity is zero.

The horizontal distance traveled (d ) is given as 3 meters and the vertical distance (h ) is 2 meters.

Using the equation of motion for vertical motion:

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

Since the initial vertical velocity is zero, the term involvingv_0 disappears.

Simplifying the equation:

2 = \frac{1}{2} \cdot 9.8 \cdot t^2

2 = 4.9 \cdot t^2

t^2 = \frac{2}{4.9}

t^2 \approx 0.4082

Taking the square root of both sides:

t \approx 0.6397 seconds

Now, using the equation of motion for horizontal motion:

d = v_0 \cdot t

Substituting the given values:

3 = v_0 \cdot 0.6397

v_0 = \frac{3}{0.6397}

v_0 \approx 4.69 m/s

Therefore, the stunt man needs to be traveling at approximately 4.69 m/s horizontally to land on the roof of the garage.

\textbf{Answer:} The stunt man needs to be traveling at approximately 4.69 m/s horizontally.

Let's assume the initial velocity of the stunt man is

The horizontal distance traveled (

Using the equation of motion for vertical motion:

Since the initial vertical velocity is zero, the term involving

Simplifying the equation:

Taking the square root of both sides:

Now, using the equation of motion for horizontal motion:

Substituting the given values:

Therefore, the stunt man needs to be traveling at approximately 4.69 m/s horizontally to land on the roof of the garage.

\textbf{Answer:} The stunt man needs to be traveling at approximately 4.69 m/s horizontally.

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