Question

An object is thrown up at 20m/s if it hits the ground after 10s,determine the final velocity and the height

Answer 1

1) The final velocity of the object is -78 m/s (downward)

2) The final height of the object is -290 m

Explanation:

1)

When the object is thrown up, it follows a free fall motion, subjected only to the force of gravity (that pulls the object downward). Therefore, the object has a constant acceleration

$g=9.8 m/s^2$

towards the ground (acceleration of gravity).

Since this is a motion with constant acceleration, we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a = g is the acceleration

t is the time of flight

In this problem we have:

u = +20 m/s (taking upward as positive direction)

t = 10 s is the time of flight

$g=-9.8 m/s^2$ (downward, so negative)

Solving for v, we find the final velocity:

$v=20+(-9.8)(10)=-78 m/s$

2)

For this part of the problem, we can use another suvat equation:

$s=ut+\frac{1}{2}at^2$

where

u is the initial velocity

a is the acceleration

t is the time

s is the displacement of the object

In this problem,

u = +20 m/s

t = 10 s

$g=-9.8 m/s^2$

Substituting, we find the final position of the object with respect to the initial position:

$s=(20)(10)+\frac{1}{2}(-9.8)(10)^2=-290 m$

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