Question #bc8e6

1 Answer
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 uu
b) the vertical velocity of the bean bag thrown at an angle is u* cos50^0ucos500
We know that acceleration is -gg, acceleration due to gravity (negative since acting in the opposite direction to the vertical velocity),
We can now use following definition of acceleration :

a=("final velocity"-"initial velocity")/ta=final velocityinitial velocityt

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

a=-9.8= "-initial velocity" /ta=9.8=-initial velocityt
and so
t="initial velocity"/9.8t=initial velocity9.8

For a) t=u/9.8t=u9.8

For b) t=u/9.8*cos50^0=u/9.8*0.64t=u9.8cos500=u9.80.64

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