a body is projected upwards with a velocity of 50 m/s from the top of a tower 100 m high.How long will it take to reach the ground? What will be the velocity with the body strikes the ground?
12 sec
To determine the time it takes for the body to reach the ground, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = displacement (distance traveled)
u = initial velocity
a = acceleration (in this case, due to gravity, which is approximately -9.8 m/s^2)
t = time
Given:
u = 50 m/s (upwards)
s = -100 m (negative because the displacement is downwards, opposite to the direction of the initial velocity)
We can rearrange the equation to solve for time:
s = ut + (1/2)at^2
-100 = (50)t + (1/2)(-9.8)t^2
Simplifying the equation:
-100 = 50t - 4.9t^2
Rearranging the equation to a quadratic form:
4.9t^2 - 50t - 100 = 0
Now we can solve this quadratic equation to find the value of 't'. Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values:
t = [ -(-50) ± √((-50)^2 - 4(4.9)(-100)) ] / (2 × 4.9)
Simplifying:
t = [ 50 ± √(2500 + 1960)] / 9.8
t = [ 50 ± √(4460)] / 9.8
t = [ 50 ± 66.83] / 9.8
This gives us two possible values:
t = (50 + 66.83) / 9.8 ≈ 11.38 seconds
t = (50 - 66.83) / 9.8 ≈ -1.72 seconds (extraneous, as time cannot be negative)
Therefore, it will take approximately 11.38 seconds for the body to reach the ground.
To find the velocity of the body when it strikes the ground, we can use another equation of motion:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
u = 50 m/s (upwards)
a = -9.8 m/s^2 (acceleration due to gravity, acting downwards)
t = 11.38 seconds (as calculated above)
Plugging in the values:
v = 50 + (-9.8)(11.38)
v = 50 - 111.524
Therefore, the velocity with which the body strikes the ground is approximately -61.524 m/s (negative sign indicates that it is moving downwards).