Moments later, your friend passes the ball back up to you. Your friend throws it straight upwards at a speed of 16.6 m/s. If you catch it when it is still travelling upward at 5.3
m/s how far apart are you and your friend?

Analysing the question:

since the ball is thrown upwards, the gravity will be applied opposite to the motion of the ball

We are given:

acceleration due to gravity (a) = -10 m/s²

initial upward velocity (u) = 16.6 m/s

final upward velocity (v) = 5.3 m/s

distance between the friends (s) = s m

Neglecting air resistance

Solving for the distance between the friends:

from the first equation of motion

v = u + at

replacing the variables

5.3 = 16.6 + (-10)(t)

-10(t) = 5.2 - 16.6

-10 (t) = -11.3

t = 1.13 seconds

from the second equation of motion

s = ut + 1/2 at²

replacing the variables

s = (16.6)(1.13) + -5(1.13)²

s = 18.8 - 6.4

s = 12.4 m

Therefore, the 2 friends are 12.4 m apart

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6

massxgravity x heght=weightxheight= pe

4900=490xh

h=10