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height displacement

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1.3 A stone is thrown at a velocity of42 m/s at an angle of26°to the horizontal. Calculate the following: 1.3.1 The maximum height that the stone reaches 1.3.2 The horizontal displacement of the stone
asked Oct 27, 2014 in PHYSICS by anonymous

2 Answers

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1.3.1)

Find the maximum height of the projectile:

H = v0² sin²θ / 2g

Here, θ is the angle with horizontal, v0 is the initial projectile velocity.

Substitute 42 m/s for v026° for θ, and 9.81 m/s² for g.

H = (42)² sin²26° / 2(9.81)

  = 17.278 m

answered Oct 27, 2014 by bradely Mentor
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1.3.2)

Find the horizontal displacement of the stone of the projectile:

R = v0² sin2θ / g

Here, θ is the angle with horizontal, v0 is the initial projectile velocity.

Substitute 42 m/s for v026° for θ, and 9.81 m/s² for g.

R = (42)² sin2(26°) / (9.81)

  = 141.697 m

answered Oct 27, 2014 by bradely Mentor

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