2. Dominic wants to hurl a water balloon through his younger brother Anthony’s o

Question

2. Dominic wants to hurl a water balloon through his younger brother Anthony’s open window using a launcher placed on the ground some distance away from the house. The window is 11 m above the point at which the balloon leaves the launcher. The launcher is aimed at 60 degrees.

a) Assuming that the water balloon is at its highest point when it passes through the window, what is the initial velocity in the y-direction?

b) At what velocity does the water balloon leave the launcher?

c) How far away from the wall of the house must the launcher be placed?

Explanation / Answer

Let the vertical component of the launch velocity = u.

Then, using the equation

v² = u² + 2as, with v = 0, a = 9.81 m/s². s = 11 m

0 = u² - 2(9.81) (11)

u² = 215.82, hence u =14.69 m/s

Answer (a) is 14.69 m/s




If the actual launch velocity = V,

vertical component = V sin 60º

so V (3/2) = 14.69

V = 14.69 (2/3) = 16.96 m/s

Answer (b) = 16.96 m/s



Using the equation v = u + at with the vertical motion,

with v = 0, u = 14.69 m/s (as found above), a = 9.81 m/s²

0 = 14.69 - 9.81 t

t = 14.69 / 9.81 = 1.50 second.


Horizontal velocity = 16.96 cos 60º = 8.48 m/s

Therefore horizontal distance to window = 8.48 x 1.50 = 12.72 m

Answer (c) : the launcher must be 12.72 m from the house.

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