Question

from the top of the tower 156.8 m high a projectile is thrown up with velocity of 39.2

Answer 1

Question:-

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From the top of a tower 156.8 m high a projectile is thrown up with a velocity of 39.2m/s, making an angle 300with the horizontal. Find the distance from the foot of the tower where it strikes the ground and the time taken by it to do so. g=9.8 m/s2

Answer:-

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horizontal --- Vx = 39.2 cos 30 = 33.95 m/3

Vertical --- initial Vy = 39.2 sin 30 = 19.6 m/s

vertical distance --- d(y) = h(0) + Vy t + (1/2)gt^2

0 = 156.8 + 19.6t - 4.9t^2

0 = -32 - 4t + t^2 = (t - 8)(t + 4) <<< t = 8 (t = -4 is not possible --- would involve time travel)

so time to hit the ground is 8 seconds <<< answer

horiz. distance = Vx t = 33.95 (8) = 271.6 m <<< answer

Answer 2

equation of motion for the projectile 

y  = x  tanФ  -  g x² Sec² Ф / (2 u²)

=>  y = x/√3   -  9.8 (4/3) x²/ (2 * 39.2²)             as  Ф  = 30°

=> y = x/√3  - x² /235.2

At  t=0,   x =0  and y = 0 

We have to find x when y = - 156.8 m

so:   x²/235.2 - 156.8 - x/√3  = 0

 =>  √3 x²  - 235.2 x - 235.2 * 156.8√3 = 0

   x = [ 235.2 + - √(235.2² + 4 * 235.2 * 156.8 * 3) ] / (2√3)

     = [ 235.2 + - 235.2 *√(1+8) ]/(2 √3)

     =   2 * 235.2 /√3  meters   or   ignoring negative value.

This is the distance from the foot...

Time taken to hit the ground =  x /(u cosФ)

  = [ 2 * 235.2/√3 ] / (39.2 * √3 / 2)  sec

  = 8 sec

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