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Jake launches a water balloon at an angle of 35° above the horizontal. If he sends it flying with an initial velocity of 3 m/s, how far away does Fred (who is the same height as Jake) need to be for it to hit him (assuming Jake has a good aim)?

Sagot :

hamzaahmeds

Answer:

R = 0.86 m

Explanation:

The formula for the range of the projectile motion can be used here:

[tex]R = \frac{v^2 Sin2\theta}{g}[/tex]

where,

R = Range of projectile = distance between Jake and Fred = ?

v = launch speed = 3 m/s

θ = Launch Angle = 35°

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]R = \frac{(3\ m/s)^2Sin[(2)(35^o)]}{9.81\ m/s^2}\\\\[/tex]

R = 0.86 m

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