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

[Quantum Mind]

Homework Statement

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.

Homework Equations

S = ut - 1/2 gt² where S is the distance, u = 40 Sin[itex]\Theta[/itex] and g is 9.8 m/s² and t=2 seconds.

S = v*t

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?

 

[Hootenanny]

Homework Statement

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.

Homework Equations

S = ut - 1/2 gt² where S is the distance, u = 40 Sin[itex]\Theta[/itex] and g is 9.8 m/s² and t=2 seconds.

S = v*t

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?

You have calculated the angle of projection. You are also given the initial speed and hence you can compute the initial vertical velocity. From that you can work out the total flight time without needing the maximal height.

 

[Quantum Mind]

I get it now, thanks.