An object is thrown vertically from a height of #15 m# at # 6 m/s#. How long will it take for the object to hit the ground?

2 Answers
Nov 7, 2017

Answer:

2.4 seconds

Explanation:

If #t_1# is the time taken by object to reach same height then

#t_1 = (2u)/g = (2 × 6 ms^-1) / (9.8 ms^-2) = 1.2 s#

Velocity of object when it hits the ground is

#v = sqrt(u^2 + 2aS)#

#v = sqrt((6 ms^-1)^2 + (2 × 9.8 ms^-2 × 15m))#

#v = 18.2 ms^-1#

If #t_2# is further time taken by object to reach the ground then

#t_2 = (v - u)/g = (18.2 ms^-1 - 6 ms^-1)/(9.8ms^-2) = 1.2 s#

Total time taken #= t_1 + t_2 = 1.2s + 1.2s = 2.4s#

Nov 12, 2017

Answer:

The time is about 2.5 s.

Explanation:

Compared to your first answer, this is an alternate method of solving.

I will choose to have up be the positive direction. Use the kinematic (suvat) formula
#s = u*t + 1/2 * a*t^2#
where

  • #s = -15 m# because it will hit the ground - below the point it was thrown from
  • #u = 6 m/s# positive because it was thrown up
  • #a = -g = -9.8 m/s^2#.

Now, plugging in the data and solving for t
#-15 m = 6 m/s*t + 1/2*(-9.8 m/s^2)*t^2#

Adjust so that all 3 are on one side and zero is on the other (preparing for use of the quadratic equation)

# 1/2*9.8 m/s^2*t^2 - 6 m/s*t -15 m = 0#

# 4.9 m/s^2*t^2 - 6 m/s*t -15 m = 0#

#t = (6 +-sqrt(6^2 - 4 * 4.9 * (-15)))/(2*4.9)#

#t = (6 +-sqrt(36 + 294))/(9.8)#

#t = (6 +-sqrt(330))/(9.8) = (6 +-18.17)/(9.8)#

The - part of the #+-# will yield a negative time, so that solution is invalid. I will continue using the + part of the #+-#.

#t = (24.17)/(9.8) = 2.47 s#

I hope this helps,
Steve