Your friend, a world-class long jumper, is trapped on the roof of a burning building. His only escape route is to jump to the roof of the next building. Fortunately for him, he is in telephone contact with you, a Physics 161 student, for advice on how to proceed. He has two options. He can jump to the next building by using the long-jump technique where he jumps at 45o to the horizontal. Or, he can take his chances by staying where he is in the hopes that the fire department will rescue him. You learn from the building engineers that the next building is 10 m away horizontally and the roof is 3 m below the roof of the burning building. You also know that his best long-jump distance is 7.9 m. What do you advise him to do

the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.

Explanation:

his problem must be solved with the missile launch equations.

Let's start by looking for the jumper's initial velocity

R = v₀² sin 2θ / g

for the long jump the angle used is tea = 45º, in the exercise they indicate that the best record is R = 7.9m

v₀² = R g / sin 2te

v₀ = [tex]\sqrt{ \frac{7.9 \ 9.8}{1 }[/tex]

v₀ = 8.80 m / s

Now suppose you jump with this speed to get to the other building, let's use trigonometry for the components of the speed

sin 45 = [tex]v_{oy}[/tex] /v₀

cos 45 = v₀ₓ / v₀

v_{oy} = v₀ sin 45

v₀ₓ = v₀ cos 45

v_{oy} = 8.8 sin 45 = 6.22 m / s

v₀ₓ = 8.8 cos 45 = 6.22 m / s

now let's calculate the sato with these speeds

x = [tex]v_{ox}[/tex] t

the minimum jump is x = 10 m

t = x / v₀ₓ

t = 10 / 6.22

t = 1.61 s

let's find the vertical distance for this time

y = v_{oy} t - ½ g t²

where zero is placed on the jump building

y = 6.22 1.61 - ½ 9.8 1.61²

y = -2.69 m

Let's analyze this result, the negative sign indicates that it is descending and the distance is less than the difference in height between the two buildings, therefore the person would be saved in the jump.