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Projectile motion
I need help trying to help one of my high school kids answer a projectile motion problem.
A bride throws a bouqet from a balcony 20 meters from the ground at an angle of 37 degrees above the horizontal with an initial speed of 10 m/s. The bouqet overshoots the crowd below and lands on the ground. How far from the balcony did the bouqet land. Do I need to know the maximum height of the bouqet's path? Casey |
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bride throws a bouqet from a balcony 20 meters from the ground at an angle of 37 degrees above the horizontal with an initial speed of 10 m/s. The bouqet overshoots the crowd below and lands on the ground. How far from the balcony did the bouqet land.
Horizontal component of velocity = 10 Cos 37 Vertical component of velocity = 10 Sin 37 Vertical acceleration = -g Vertical distance = -20 Horizontal acceleration = 0 Find the time taken to fall to the ground (using only vertical component) (Formula: s = ut + 1/2 ft^2, where u is initial velocity, f is acceleration, t tis time and s is distance) Multiply this time by horizontal velocity For maximum height, this occurs when vertical velocity = 0 You can use v^2 - u^2 = 2fs for this |
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