A Year 12 student tossed a beanbag with an initial velocity ‘u’ at a launch angle of 35.00 from thehorizontal. It struck the circular target located 15.5 m away from the bag and 0.200 m above theground. (See diagram below)Figure 1: A student played a beanbag toss game (not to scale)Determine the magnitude of the initial velocity ‘u’ of the beanbag. Ignore air resistance.

To solve this problem, we can use the equations of motion under constant acceleration. In this case, the acceleration is due to gravity and acts vertically downwards.

First, we can split the initial velocity into horizontal and vertical components. The horizontal component (u_x) is given by ucos(35) and the vertical component (u_y) is given by usin(35).

The horizontal motion is uniform, so the horizontal distance covered (15.5 m) is given by the horizontal speed (u_x) times the time of flight (t). This gives us our first equation:

15.5 = u*cos(35)*t ----(1)

The vertical motion is under constant acceleration due to gravity. The vertical displacement (0.2 m) is given by the initial vertical speed (u_y) times the time of flight (t) minus half the acceleration due to gravity (g) times the square of the time of flight. This gives us our second equation:

0.2 = usin(35)t - 0.5gt^2 ----(2)

We can solve these two equations simultaneously to find the initial speed u.

First, we can solve equation (1) for t:

t = 15.5 / (u*cos(35))

Substitute this into equation (2):

0.2 = usin(35)(15.5 / (ucos(35))) - 0.5g*(15.5 / (u*cos(35)))^2

Solving this equation for u gives the magnitude of the initial velocity.

Note: The value of g (acceleration due to gravity) is approximately 9.8 m/s^2.