It rises till all of its Kinetic energy is converted into potential energy.
so, mgh=(1/2)m(v^2)
so, h=(v^2)/2g = 12*12/(2*9.81)=7.34 m
Answer:
Explanation:
My speed after the interaction will depend upon the impulse the ball will make on me . Now impulse can be expressed as follows
Impulse = change in momentum
change in momentum in the ball will be maximum when the ball bounces back with the same velocity which can be shown as follows
change in momentum = mv - ( - mv ) = 2mv
So when ball is bounced back with same velocity , it suffers greatest impulse from my hand . In return , it reacts with the same impulse on my hand pushing me with greatest impulse according to third law of motion. this maximizes my speed after the interaction.
Answer:
3 x 10^5 J
Explanation:
mass of substance, m = 1 g = 0.001 kg
Velocity of light, c = 3 x 10^8 m/s
According to the Einstein mass energy equivalence, the energy associated with the mass is given by
E = m c^2
E = 0.001 x 3 x 10^8
E = 3 x 10^5 J
Answer:
22m/s
Explanation:
To find the velocity we employ the equation of free fall: v²=u²+2gh
where u is initial velocity, g is acceleration due to gravity h is the height, v is the velocity the moment it hits the ground, taking the direction towards gravity as positive.
Substituting for the values in the question we get:
v²=2×9.8m/s²×25m
v²=490m²/s²
v=22.14m/s which can be approximated to 22m/s
I would said A is the best option if i’m wrong sorry
