Question

An object is thrown vertically from a height of `7` `m` at `4` `ms^-1`. How long will it take for the object to hit the ground?

Answer 1

Assuming the object has initial upward velocity `->t =1.67` `s`
Assuming the object has initial downward velocity `->t =0.85` `s`

Explanation:

We have been given the following informations

The object is thrown *with velocity `u=4` `ms^-1`
from a height of `h= 7` `m` and if it takes time t sec to reach the ground then to apply the equation of motion we may go as follows

Assuming the object has initial upward velocity

• `u =+4` `ms^-1-> "upward taken +ve"`
• `h= -7` `m ->"downward taken -ve"`
• Acceleration due to gravity `g=-9.8` `ms^-2 ->"downward taken -ve"`

So the equation, `h=ut+1/2g t^2` becomes
`=>-7=4xxt-1/2xx9.8t^2`
`=>4.9t^2-4t-7=0`

`t=(-(-4)+sqrt((-4)^2-4xx4.9(-7)))/(2xx4.9)=1.67s` [negative value of t neglected]

Assuming the object has initial downward velocity

• `u =+3` `ms^-1-> "downward taken +ve"`
• `h= +7` `m ->"downward taken +ve"`
• Acceleration due to gravity `g=+9.8m/s^2 ->"downward taken +ve"`

So the equation,`h=ut+1/2g t^2` becomes
`=>7=4xxt+1/2xx9.8t^2`
`=>4.9t^2+4t-7=0`

`t=(-4+sqrt((4)^2-4xx4.9(-7)))/(2xx4.9)=0.85s` [negative value of t neglected]