A body is thrown vertically upward with initial velocity 9.8m/s what height it will attain ?
H=Vi^2/2g solve this and u will get answer
yes
(1/2) m v^2 = m g h
so
h = (1/2) v^2/g
4.9m and ur creact ans give
To find the height the body will attain when thrown vertically upward with an initial velocity of 9.8 m/s, we can use the equation for the motion of a body in free fall:
h = (v^2 - u^2) / (2g)
Where:
h is the height attained by the body,
v is the final velocity (which will be zero at the highest point),
u is the initial velocity,
g is the acceleration due to gravity.
Given:
u = 9.8 m/s (initial velocity)
g = 9.8 m/s^2 (acceleration due to gravity)
Plugging the given values into the equation:
h = (0^2 - 9.8^2) / (2 * 9.8)
Simplifying:
h = (-96.04) / 19.6
h ≈ -4.9 meters
Since height cannot be negative in this context, the negative sign indicates the direction (downward) of the initial velocity. Thus, the height attained by the body when thrown vertically upward with an initial velocity of 9.8 m/s is approximately 4.9 meters.