a stone is thrown vertically upward wit a speed of 20 m/s ?

Question

a. find itsz speed after 1 sec; 2 secs
b. how far up does it travel
c. how long will it take to get back to the original time

the formula'sz vf = vi + (-g) t

Answer 1

A stone is thrown vertically upward wit a speed of 20 m/s ?
a. find its speed after 1 sec; 2 secs

v = u + at is the formula to be used. Let us assume acceleration due to gravity g as 10 m/sec^2 (exact value is 9.81 m/sec^2)

v = 20 - 10.1 = 10 m/sec after 1 sec.
v = 20 - 10.2 = 0 m/sec after 2 secs.

b. how far up does it travel

v^2 - u^2 = 2as is the formula. v is zero when the stone reaches the max. height

0 - 20 ^2 = - 2 .10. s where s is the distance.

s = 20 x 20 / 20 = 20 metres

c. how long will it take to get back to the original time

You mean how long will it take to get back to the ground? Same time , 2 seconds.

Answer 2

a) so final velocity (vf) = initial velocity (vi) plus time elapsed (t) multiplied by acceleration due to gravity in the direction of the initial velocity (-g).
vf(1 second)=20 -9.81 , vf(2 seconds) = 20 - 2*9.81

b) distance travelled is integral of velocity over the relevant time so draw yourself a velocity time graph and calculate the area under the straight line.

c) it will never get back to the original time. - perhaps you mean original height?

Answer 3

u = 20m/s
v = 0
s = ?
t = 1s
g = 9.8m/s^2
distance travelled after 1 sec is:
s= ut+1/2 gt^2
s= 20*1 + 1/2*9.8*1
= 20+4.9 = 24.9m
the speed = v = s/t
v = 24.9/1 = 24.9m/s

someone check dis..i think its wrong!!

Answer 4

for free fall:
y=y0+v0*t+1/2g*t^2
v=v0+g*t
g=-9.8m/s2
a. t=1s, v=20-9.8*1=10.2 m/s
t=2s, v=20-9.8*2=0.4 m/s

b. v=0 in up position then: h=0+20*t-1/2*(-9.8)*t^2 and
h=1/2*(-9.8)*t
solving you get: [h = 0 ∧ t = 0, h = 24.9 ∧ t = -5.081632653]
then h=24.9 m

c. h=24.9 then solving 24.9=0+20*t-1/2*(-9.8)*t^2 we get
t = -5.081632653 ∨ t = 1
for take back: t=2s