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