A projectile is shot at a velocity of $35 m/s$ and an angle of $pi/6$ . What is the projectile's peak height?

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A projectile is shot at a velocity of $35 m/s$ and an angle of $pi/6$ . What is the projectile's peak height?

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Answer 1 · 2016-02-02T19:06:13

15.6 m

Explanation:

See this URL https://en.m.wikipedia.org/wiki/Projectile_motion

Equation for the peak height is
H = $v^2sin^2 (theta)/(2g)$

This is derived using the fundamental equations of motion and the assumption that the trajectory is parabolic. The objects vertical velocity at the peak height is zero m/s.

v = 35 m/s
Theta =30 degrees
g=9.8m/s^2 Gravitational acceleration

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A projectile is shot at a velocity of $35 m/s$ and an angle of $pi/6$ . What is the projectile's peak height?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

A projectile is shot at a velocity of 35 m/s 35 m/s and an angle of pi/6 pi/6 . What is the projectile's peak height?

1 Answer

Explanation:

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Impact of this question

A projectile is shot at a velocity of $35 m/s$ and an angle of $pi/6$ . What is the projectile's peak height?