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

Answer:

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!!!