Question

if theta is the angle of projection is the range which is the maximum height time of flight then tan theta equalsand maximum height equals to

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are the horizontal range and maximum height equal in projectile motion?

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Roy Narten, Mechanical Engineer

Answered Jan 5, 2017

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Always watch your sign convention when using Newton’s equations of motion. I always assume positive = up and negative = down, so the acceleration of gravity is *always* negative:

Let’s start with the VERTICAL MOTION:

Use this equation of motion to find the time to reach the maximum height:

I like to add subscripts to help:

Since  and at maximum height, 

time to reach maximum height is 

To find the maximum height we write another familiar equation of motion using subscripts:

substitute in the time to reach max height gives

or

or maximum height is 

Now consider the HORIZONTAL MOTION:

Writing the same equation of motion except using subscripts for motion in the x-direction gives:

We neglect air resistance, so 

The time to reach maximum horizontal range will be twice the time to reach maximum height because the time to go up is exactly the same as the time to come back down.

Substituting in  and 

gives 

To answer your question, we equate 

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Jatin Rajput, Quantum Mechanics is not flawed. If it went on strike, we would reach dark ages.

Answered Apr 16, 2016

Horizontal Range:

(U^2 Sin 2 (A)/g) where 'A' is the projection angle.

Maximum Height:

U^2 [Sin (A)]^2/2 g where 'U' is the projection velocity.

[Sin (A)] ^2 = 2× Sin 2 (A)

Sin (A) = 4 × Cos (A)

tan [A] = 4, A = tan~[4] where '~' is the inverse.

· View Upvoters · Answer requested by Bindu Bindu

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