A body is thrown vertically upwards having the velocity of 10m/s. It's velocity when it touches the ground is
same as when it left.
momentum is conserved.
actually, its speed is the same. The velocity has opposite direction.
To find the velocity of the body when it touches the ground, we need to consider the motion of the body.
When a body is thrown vertically upwards, it experiences a constant acceleration due to gravity acting in the opposite direction to its motion. The acceleration due to gravity on the surface of the Earth is approximately 9.8 m/s² (downwards).
Given that the initial velocity of the body is 10 m/s upwards, we can find the time it takes for the body to reach its peak height using the equation of motion:
v = u + at
where:
- v is the final velocity (0 m/s at the peak height)
- u is the initial velocity (10 m/s upwards)
- a is the acceleration due to gravity (-9.8 m/s²)
- t is the time
Rearranging the equation, we have:
t = (v - u) / a
t = (0 - 10) / -9.8
t = 1.02 seconds (approximately)
Now, we can find the velocity of the body when it touches the ground by considering the total time of the motion. Since the time taken to reach the peak height is 1.02 seconds, the total time of the motion (upwards and downwards) will be twice that:
Total time = 2 * 1.02 = 2.04 seconds
To find the velocity when it touches the ground, we can use the equation of motion:
v = u + at
where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration due to gravity
- t is the total time
Plugging in the values, we get:
v = 10 + (-9.8) * 2.04
v = 10 - 19.92
v = -9.92 m/s
Therefore, the velocity of the body when it touches the ground is approximately -9.92 m/s. The negative sign indicates that the velocity is in the opposite direction as the initial velocity (downwards).