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During a medieval siege of a castle, the attacking army uses a trebuchet to hurl heavy stones at the castle walls. If the trebuchet launches the stones with a velocity of +30.0 m/s at an angle of 50.0°, how long does it take the stone to hit the ground? What is the maximum distance that the trebuchet can be from the castle wall to be in range? How high will the stones go? Show all your work.

Respuesta :

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Answer:The above given quantities can be calculated in following steps;

Explanation:

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The projectile reaches a maximum height of 26.4 m.

The question is demanding us to find the;

i) Time of flight:

T = 2usinθ/g

u = initial velocity

θ = angle of projection

g = acceleration due to gravity

T = 2 × 30.0 × sin 50.0°/10

T = 4.6 s

ii) The range;

R = u^2sin2θ/g

R = (30.0)^2 × sin 2( sin 50.0°)/10

R = 88.6 m

ii) The maximum height;

H = u^2sin^2θ/2g

H =  (30.0)^2 × sin^2(50.0°)/2 × 10

H = 26.4 m

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