(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A projectile is launched with an velocity of 40 m/s and crosses a tower of height 20.4 m after 2 seconds. Find the angle of projection with the horizontal and the horizontal range.

2. Relevant equations

S = ut - 1/2 gt^{2}where S is the distance, u = 40 Sin[itex]\Theta[/itex] and g is 9.8 m/s^{2}and t=2 seconds.

S = v*t

3. The attempt at a solution

I have found the angle of projection which is 30 degrees, but I am not getting the horizontal range. The answer is 141.3 m while I keep getting 138.56 (I used the equation distance = velocity * time i.e. 40Cos[itex]\Theta[/itex] * 4 ).

As I see it, the difficulty is that the maximum height reached is not known as the problem states that the projectile crosses the tower in 2 seconds and it doesn't say that it is the maximum height reached. Therefore the time t for the total time traveled cannot be equal to 4 seconds. How to proceed further?

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# Finding the horizontal range of a projectile when maximum height is not known

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