How to interpret this projectile motion problem

[mcintyre_ie]

Hey
I just need some help with intepreting this question. I've allready solved the first part (A), but I am not sure as to when the time t1 occurs. Heres the question:
(A) A particle is progected with speed u at an angle alpha to the horizontal. If the maximum height reached is the same as the total horizontal range, show that tan alpha = 4
(B) The particle moves at right angles to its original direction of motion after a time t1 and then strikes the horizontal plane after 8s, both times measured from the instant of projection.
Show u = g(root)17
Calculate t1

Ok, so I've shown tan alpha = 4 and u = g(root)17, but i don't know when t1 occurs. Is it when the maximum vertical height is achieved? Any help is appreciated.
Thanks in advance

 

[HallsofIvy]

Draw a simple picture: a horizontal straight line and a second line at angle α to it. That second line represents the initial direction of velocity. Draw its horizontal and vertical components. They have length, of course, of u cosα and u sinα resepectively. Now at the end of the second line draw a third line perpendicular to the second. Also draw a vertical line from the point of intersection of second and third and complete the right triangle with a horizontal line.

That represents the velocity vector, with horizontal and vertical components, when "the projectile is moving at right angles to the initial trajectory". What is the angle at the top of this right triangle? You should be able to see that the two right triangles are congruent and comparing the horizontal and vertical components, that the horizontal component of velocity is now v sinα and the vertical component is -v cosα (v is the speed at t₁. It is not the same as u!). You know that v_x= u cosα and that -v cosα= u sinα- g t₁. From those two equations you can find v and t₁.

 

[M98Ranger]

To interpret this projectile motion problem, you first need to understand the basic principles of projectile motion. In this problem, a particle is projected at an angle alpha to the horizontal with an initial speed of u. The maximum height reached by the particle is equal to the horizontal range, which means that the particle will reach the same height at a certain point in its trajectory as it does when it reaches the end of its horizontal range. This point is known as the maximum height or vertex of the trajectory.

In part (A), you have already solved for the value of tan alpha, which is equal to 4. This means that the angle alpha is equal to tan^-1(4) or approximately 75.96 degrees. This angle is important because it helps determine the trajectory of the particle.

In part (B), it is stated that after a time t1, the particle moves at right angles to its original direction of motion and then strikes the horizontal plane after 8 seconds. This means that after t1, the particle changes its direction of motion from being horizontal to moving vertically downward towards the ground. This occurs because the particle has reached its maximum height and begins to fall back down towards the ground.

To calculate t1, you will need to use the equations of motion for projectile motion. You have already solved for u in part (A), which is equal to g(root)17. Using this value for u, you can then use the equation t1 = (2u*sin alpha)/g to solve for t1. Once you have calculated t1, you can use it to determine the time at which the particle changes its direction of motion and begins to fall towards the ground.

In summary, interpreting this projectile motion problem involves understanding the concepts of maximum height, angle of projection, and equations of motion. By solving for the given values and using the equations of motion, you can determine the time at which the particle changes its direction of motion and calculate t1.