# An object is thrown vertically from a height of #9 m# at #43 m/s#. How long will it take for the object to hit the ground?

##### 1 Answer

#### Answer:

#### Explanation:

The fall time and rise time will be different as the object begins and ends its motion at different altitudes. Ignoring air resistance, you can use kinematic equations to calculate the total flight time.

For the rise time:

#v_f=v_i+aDeltat_(rise)#

The object is launched at an initial velocity (

Solve for

#Deltat_(rise)=(cancel(v_f)-v_i)/a#

#Deltat_(rise)=(-43m/s)/(-9.8m/s^2)=4.4s#

If the object began and ended at the same altitude, we could simply multiply this result by two to get the total flight time. However, the object begins at

#(v_f)^2=(v_i)^2+2aDeltay#

#Deltay=(-(v_i)^2)/(2a)#

#Deltay=(-(43m/s)^2)/(2*-9.8m/s^2)#

#Deltay=94.34m#

This is the maximum altitude of the object. Adding *begins* from its maximum altitude this time.

#Deltay=cancel(v_iDeltat_(fall))+1/2a(Deltat_(fall))^2# Solve for

#Deltat_(fall)#

#Deltat_(fall)=sqrt((2Deltay)/a)#

#Deltat_(fall)=sqrt((2(-(94.34+9))m)/(-9.8m/s^2))#

#Deltat_(fall)=4.59s#

We can add these times together to get the total flight time:

#t_(t o t)=t_(fall)+t_(rise)=4.59s+4.4s=8.99s#

This is