An object is thrown vertically from a height of #7 m# at # 3 m/s#. How long will it take for the object to hit the ground?

1 Answer
Jun 20, 2017

Answer:

#t = 1.54# #"s"#

Explanation:

We're asked to find the time #t# when the object hits the ground, given that it was thrown vertically upward with a speed of #3"m"/"s"#, at a height of #7# #"m"#.

As soon as the object is let go, it is in a state of free-fall, and the kinematics equations can predict its motion, such as its maximum height reached, time of flight, etc.

To find this time, we can use the equation

#y = y_0 + v_(0y)t + 1/2a_yt^2#

For this equation,

  • The height #y# is ground level, which I'll call #0# #"m"#

  • The initial #y_0# is #7# #"m"#

  • The initial #y#-velocity #v_(0y)# is #3"m"/"s"#

  • The time #t# is when it is at height #y = 0#, which is what we're trying to find

  • The #y#-acceleration #a_y# is equal to #-g#, which is #-9.8"m"/("s"^2)#

Plugging in our known values, we have

#0# #"m" = 7# #"m"# #+ (3"m"/"s")t + 1/2(-9.8"m"/("s"^2))t^2#

#(-4.9"m"/("s"^2))t^2 + (3"m"/"s")t + 7# #"m"# #= 0#

Using the quadratic formula:

#t = (-3+-sqrt((3)^2 - 4(-4.9)(7)))/(2(-4.9)) = color(red)(1.54# #color(red)("s"# (positive solution)

Thus, the object will strike the ground after #color(red)(1.54# #sfcolor(red)("seconds"#.