An object is thrown vertically from a height of 9 m at 6 m/s. How long will it take for the object to hit the ground?

2 Answers
Mar 29, 2018

The time to hit the ground is =2.099s

Explanation:

Resolve in the vertical direction uarr^+

The initial velocity is u=6ms^-1

The acceleration due to gravity is g=-9.8ms^-2

The distance is s=-9m

Applying the equation of motion

s=ut+1/2at^2

-9=6t-9.8/2t^2

4.9t^2-6t-9=0

Solving this quadratic equation in t

t=((6)+-sqrt(36+4*9*4.9))/(2*4.9)

=(6+-sqrt(212.4))/9.8

t_1=(6+14.57)/9.8=2.099s

t_2=(6-14.57)/9.8=-0.87s

The time to hit the ground is =2.099s

graph{4.9x^2-6x-9 [-9.99, 10.02, -5, 4.995]}

Mar 29, 2018

First, calculate the time the object takes to reach the maximum height.

To calculate the time taken to reach the maximum height, you have the following info:

color(teal)(u=6m/s, v=0,a=-9.8m/s^2 (a=-g, as the object here move upwards.)

According to the first equation of motion,

color(teal)(v=u+at.

Plug in the values:
color(teal)(0=6-9.8t
or, color(teal)(9.8t=6
or, color(teal)(t=6/9.8sapprox0.61s

To reach the same height from where it was thrown (i. e., 9m above the ground), the object will take the same time.

Thus, time taken to go up and come down to the 9m level, total time taken
color(teal)(=0.61s+0.61s=1.22s

Now, all you need is the time taken by the object to go down the last color(blue)(9m

For that,
color(blue)(S=9m,u=6m/s,a=+9.8m/s^2

Use the third equation to get:
color(blue)(v^2=u^2+2aS

Plug in the values; color(blue)(vapprox14.57

Now, find the time using the first equation of motion:
color(blue)(14.57=6+9.8t
color(blue)(tapprox0.87

Thus, total time,
color(green)(T=(0.87+1.22)s=2.09s

hope that helped...Thanks for reading!!!