Question

An object is thrown upward from the edge of a tall building with a velocity of 10 m/s. Where will the object be 3 s after it is thrown? Take

Answer 1

Do you have the height of the building?

Answer 2

the formula is\[v=u- g t\]

Answer 3

for maximum height final velocity v=0

Answer 4

\[0=10- 10 t\]

Answer 5

t=1 sec

Answer 6

so time required to reach its maximum height is 1 sec

Answer 7

it will come back to the top of the building after another 1 sec

Answer 8

so total

Answer 9

time needed to come back to the top of the building is 2 sec

Answer 10

ur throughing object will fall more in another 1 sec if the building is very high

Answer 11

it will be 20m (9.8*2), after reaching its max height (t=1)

Answer 12

20m/s

Answer 13

correct shamim?

Answer 14

no

Answer 15

u know the equation for maximum height is\[v ^{2}=u ^{2}- 2 g h\]

Answer 16

u know for maximum height final velocity v=0

Answer 17

so the equation will b\[0^{2}=10^{2}-2 \times 10 \times h\]

Answer 18

anyway i m taking gravitational acceleration g=10m/s

Answer 19

h=5m

Answer 20

maximum height h=5m

Answer 21

ya u r right @dgamma3

Answer 22

u were telling the final velocity after 3 sec

Answer 23

understood

Answer 24

there is an other way (equation)
s=Vo*t -1/2*g*t^2
Vo=10
t=3
g=10 (9.8)
solve
s= -15m (that means 15 m below point of throw)

Answer 25

ya u r right @mos