Question

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Wild E. Coyote fires an Acme rocket with an initial velocity of 35.0 m/s at an angle of 38 degree above the horizontal ground. Find the following: The rocket's time in the air The rocket's maximum height The distance the rocket travels horizontally

Answer 1

Initial velocity, at an angle $\theta= 38^\circ$ above horizontal

then Its horizontal component, $u_x= ucos\theta= (35) cos38^\circ= 27.58 m/s$

its vertical component, $u_y= usin\theta= (35) sin38^\circ= 21.548 m/s$

a). We assume air firction to be negligible, then motion will be completly uniform and thus

its time of ascent= time of descent=t

and Total time of flight or Total time in air will be, T= t+t

T=2t............(1)

Now considering its vertical motion, at highest point its final velocity will be zero.

Hence, v_f=0m/s

its vertical component of initial velocity,

acceleration, a_y= -g=-9.8m/s² ( As gravity is against its vertical motion)

time= t=T/2

Now using equation of motion,

$0= (21.548)+(-9.8)(\frac{T}{2})$ (using equation 1)

$T= \frac{21.548 \times 2}{9.8}= 4.397 s$ (ANS)

b). Again considering its vertical motion and assuming that it maximum height is H, then using equations of motion,

  

  (ANS)

c). Also In time "T", max horizontal distance projectile will cover without air friction is,

Horizontal range= ( Horizontal component of velocity)(Time of flight) (Since no acceleration acts horizontally)

$R= u_xT= 27.58 \times 4.397= 121.269 m$ (ANS)