# A projectile is shot from the ground at a velocity of #52 m/s# and at an angle of #(pi)/2#. How long will it take for the projectile to land?

##### 1 Answer

#### Explanation:

We are given the initial velocity of the projectile as well as the launch angle. We assume air resistance is negligible, so the only acceleration experienced by a simple projectile is that of gravity, which acts only in the vertical direction (i.e. downwards).

We would typically break the launch velocity up into parallel and perpendicular components given the launch angle, but this particular launch angle is

We use kinematics. We have the acceleration

Because the projectile will momentarily stop at its maximum altitude before falling, we say that the final velocity for the launch is

#v_f=v_i+a_yDeltat# Where

#v_i# is the initial velocity,#v_f# is the final velocity,#a_y# is the acceleration, and#Deltat# is the rise time.

We first solve for

#Deltat=(v_f-v_i)/(a_y)#

We know

Using our known values:

#Deltat=(-52m/s)/(-9.8m/s^2)=5.3s#

This is the time elapsed as the projectile goes from launch site to maximum altitude. The fall time will be the same, so we multiply by two.

#t_(t o t)=2xxt_(rise)=2*5.3=10.6s#

Thus, it will take

--

Note that the rise and fall times are equal only if the projectile lands at the same height is was launched from.