Question

A man stands on the roof of a building of height 14.2 m and throws a rock with a velocity of magnitude 26.9 m/s at an angle of 25.6 ∘ above the horizontal. You can ignore air resistance.

A.) Calculate the maximum height above the roof reached by the rock.

B.) Calculate the magnitude of the velocity of the rock just before it strikes the ground.

C.) Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

Answer 1

This question will be done in 2 parts. Firstly we'll need to find the maximum height. After that only we'll be able to find the range of projectile as we have fired it from a certain height.

Now For Maximum Height,

Initial Horizontal speed of projectile,

Initial Vertical speed of projectile,

Considering vertical motion only, we know that at peak, projectile comes to rest momentarily, thus at maximum height final velocity of projectile will be

Acceleration acting vertically, a_y= -g= -9.8 m/s² ( negative sign is used since gravity opposes the initial vertical motion)

Let Maximum height from point of firing be h' , then by using equations of motion,

  

  

( from initial point of firing of projectile)

then Total maximum height from ground level is H= (14.2+6.88)m= 21.08m (ANS)

Let time taken to reach that much height be t', then again using equations of motion,

  

  

  

Then horizontal distance travelled in this much time is

  ............................(1)

( As no acceleration will act on it horizontally)

Now considering 2nd part of its trip, when the projectile is at a height H= 21.08m

for this initial vertical velocity is v₁= u'= 0 m/s

vertical acceleration acting, a'= g= 9.8m/s² ( it will be positive as gravity will support this motion now)

Then final vertical velocity be v_y, then using equations of motion,

  

  

Final horizontal velocity, (SInce no acceleration acts horizontally)

then Final velocity just before it strikes is

(ANS)

Let time taken by projectile to reach ground from this much is t, then by equations of motion,

  

  

During all these motion, horizontal speed will remain same as no acceleration will act on it horizontally thus horizontal distance covered in this trip is

  .......................(2)

Then Total Range of rock is

Using equation 1 and 2 in above, we get

  ( ANS)