Answer:
After 1 sec = 4.9 m
After 2 sec = 19.6 m
After 3 sec = 44.1 m
After 4 sec = 78.4 m
After 5 sec = 122.5 m
Explanation:
After 1 sec:
<em>u=0m/s t=1 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(1) + (1/2)(9.8)(1²) = 4.9m
After 2 sec:
<em>u=0m/s t=2 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(2) + (1/2)(9.8)(2²) = 19.6m
After 3 sec:
<em>u=0m/s t=3 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(3) + (1/2)(9.8)(3²) = 44.1m
After 4 sec:
<em>u=0m/s t=4 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(4) + (1/2)(9.8)(4²) = 78.4m
After 5 sec:
<em>u=0m/s t=5 s a=9.8m/s²</em>
s = ut + (1/2)at²
=0(5) + (1/2)(9.8)(5²) = 122.5m
Answer:

Explanation:
It is given that,
Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.


Initial speed of the projectile is v and final speed is 0.5 v.


g is the acceleration due to gravity
Let h is the height above the ground. Using the second equation of motion as :



So, the height of the projectile above the ground is
. Hence, this is the required solution.
During the phase transition vapour --> liquid water, the temperature of the water does not change; the molecules of water release heat and the amounf of heat released is equal to

where
m is the mass of the water

is the latent heat of evaporation.
For water, the latent heat of evaporation is

, while the mass of the water is

so, the amount of heat released in the process is
Answer:

Explanation:
Mass of the helium gas filled inside the volume of balloon is given as




now total mass of balloon + helium inside balloon is given as


now we know that total weight of balloon + cargo = buoyancy force on the balloon
so we will have




Answer:
The answer is 15000kgms^-2
Explanation:
Actually here we are using the formula F=ma