Question

you jump from a 20 meter tall building to a 12 meter tall building you jump horozontally with a speed of 5.0 m/s.

all must be answered using ther following equations v = dx/dt;   a = dv/dt;    v = v₀ + at;    x = x₀ + v_ot + at²/2;    v² = v_o² + 2aD

how long in seconds are you in the air?

how far apart are the two buildings assuming you jumped from the edge and landed on the edge?

what is your speed when you land?

for this question you must use the pythagrium therm

Answer 1

consider vertical motion:

Vo= 0

a = -9.8 m/s^2

use:

X = Xo +Vo*t+0.5*a*t^2

12= 20 + 0*t + 0.5*(-9.8)*t^2

-8 = -4.9*t^2

t= 1.28 s

Answer: He was in air for 1.28 s

At the end vertical velocity, V=Vo+a*t

=0 - 9.8*1.28

= 12.544 m/s

--------------------------------------------------------------------------------------------------------------------------------------------------

Conider horizontal motion:

There is no acceleration in this horizontal direction.

here v = 5 m/s

v=dx/dt

5 = dx/1.28

dx = 1.28*5 = 6.4 m

Answer: Two buildings are 6.4 m apart

Final horizontalvelocity will be the same that is 5 m/s

---------------------------------------------------------------------------------------------------------------

We have,

final velocity in vertical direction = 12.544 m/s

final velocity in horizontal direction = 5 m/s

Net final velocity can be calculated by pythagoras theorem.

final velocity= sqrt(12.544^2 + 5^2) =13.5 m/s

Answer: Speed when landed is 13.5 m/s