A projectile is launched from the top of a 20 m tall building (r(0) - (0,20)m) w

Question

A projectile is launched from the top of a 20 m tall building (r(0) - (0,20)m) with an initial velocity of 10 Square root 2 m/s at an angle of 45 Degree. Assume that air resistance is negligible, and that the acceleration due to gravity is 10 m/s^3. Determine an expression for the velocity and displacement of the projectile as a function of time, f (measured in seconds). Determine the speed of the object 1 second after launch. Determine the angle between the object's velocity and acceleration vectors, 1 second after launch. How far from the base of the building will the projectile land?

Explanation / Answer

Given values:

Initial velocity = Vox = 10 m/s

= 45o

Height of the bulding = 20 m

Now,

         So xo = 0 and yo = 0

        herefore,

X = Vox t =Vocoso * t

= 10 *(1/)* t

= 10 t……………………………………………………………….equation 1

Velocity is,

Vx = Vocos =10 *(1/) = 10 m/s

x-component remains constant. Only y component changes like free fall ( vertical direction)

Vy = Vo sin – gt = 10 – 10t m/s ……………………….equation 2

   = 10 m/s

Tan = R/V

=Tan-1 (R/V) Where R = maximum range and

R = (Vo2sin2)/g = (10)2*sin(90)= 200 m

V is velocity of the object V= (2gh)(1/2)= 2*10*20 = 400

Therefore,

=Tan-1 (R/V)= Tan-1 (200/400)= Tan-1(0.5)= 26.5o

=26.5o

So,

                 R = (Vo2sin2)/g

                = (10)2*sin(90)

                  = 200 m

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