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

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