# Question #bc8e6

Mar 3, 2016

The beanbag that is thrown at an angle.....

#### Explanation:

Firstly, we need to consider velocity in the vertical direction. If we define u as the initial speed of both bean bags, then
a) the vertical velocity of the bean bag thrown straight up is $u$
b) the vertical velocity of the bean bag thrown at an angle is $u \cdot \cos {50}^{0}$
We know that acceleration is $- g$, acceleration due to gravity (negative since acting in the opposite direction to the vertical velocity),
We can now use following definition of acceleration :

$a = \frac{\text{final velocity"-"initial velocity}}{t}$

At the maximum height, the final vertical velocity will be nil, so

$a = - 9.8 = \frac{\text{-initial velocity}}{t}$
and so
$t = \frac{\text{initial velocity}}{9.8}$

For a) $t = \frac{u}{9.8}$

For b) $t = \frac{u}{9.8} \cdot \cos {50}^{0} = \frac{u}{9.8} \cdot 0.64$

So (b), the bean bag thrown at an angle , will have a shorter flight time and will hit the ground first