Question

From the top of a tower 156.8m high a projectile is throwing up with a velocity of 39.2m/sec making an angle Ф=30 with horizontal. Find the distance from the foot of the tower where it strikes the ground and the time taken to do so...........please answer it ASAP........

Answer 1

Answer:

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