Question

1)A ball is thrown vertically upwards with a velocity of 49m/s. Calculate:
a) The maximum height to which it rises
b) The total time it takes to return to the surface of Earth

2) A stone is released from a top of a tower of height 19.6m. Calculate its final velocity just before touching the ground. [G=9.8N]

Answer 1

a. u= 0m/s, v=49m/s a= -9.8m/s2
h = v2 -u2/ 2g
(49m/s)2- (0m/s)2/2* -9.8
=122.5m
b. t= v-u/a
=49-0m/s/ 9.8
=5 second




2. u=0m/s , h=19.6m a= 9.8m/s2
v2-u2=2gh
v2 = 2*9.8*19.6m
v2= 384.16m2 /s2
v=19.6m/s


Mark it as brainlist

Answer 2

1)
v²=u²-2gh
(vertically upwards so, g=-9.8m/s)
0=49²-2(9.8)(h)
0=2401-19.6h
h=2401/19.6
a) h=122.65
for t
v=u+gt
0=49-9.8t
9.8t=49
t=49/9.8
b) t=5 sec to reach the top so 10 sec to reach the surface of the earth.
2)
h=19.6
u=0
v=?
v²=u²+2gh
v²=0+2(9.8)(19.6)
v²=384.16
v=√384.16
v=19.6m/s
there are 3 formulas.
$<b> 1)S=ut+½at² 2)v=u+at 3)v²=u²+2as$
hope it helps.... ;)