Find the maximum height and the time of flight of a body fired vertically up from the ground given that it loses 60 percent of its initial speed in rising 4.2m

Answer 1

Aaz=Emanuel?

v = Vi - g t
h = .5 g t^2

at 4.2 meters, t=time at 4.2m
4.2 = .5 g t^2
t = sqrt (8.4/g)
v = Vi - sqrt (8.4 g) = 0.6 Vi
so
0.4 Vi = sqrt(8.4 g)
Vi = sqrt(8.4 g) / 0.4
use 9.81 or whatever you use for g and you have the initial speed up
0 = Vi - g t finds t at top
double that for total time in air
h = .5 g t^2 using t at top

Answer 2

At 4.2 [m], it loses 60% of its initial speed, u. This means that at 4.2 [m], the speed, v = u - 0.6u.

So, by using v^2 = u^2 - 2gs, assume g = 9.81 [m/s^2], s = 4.2 [m]

(u - 0.6u)^2 = u^2 - 2gs

u = 9.9045 [m/s], this is the initial speed.

Then, at the maximum height, v = 0. So from v = u - gt, at the maximum height,

0 = u - gt, substitute initial speed,u, and g = 9.81 [m/s^2]

the time at the maximum height, t = 1.0096 [s]. For the time of flight of a body fired vertically up from the ground is the double value of the time at the maximum height.

So, the time of flight of a body fired vertically up from the ground is 1.0096 x 2 = 2.0193 [s]

Then, to find the maximum height, by using s = ut - 0.5gt^2, where t = 1.0096 [s],

s = ut - 0.5gt^2 = 9.9045(1.0096) - 0.5(9.81)(1.0096)^2
= 9.9996 - 4.9996 = 5 [m]