a steel ball dropped from a tower strikes the ground in 3.0 seconds. Find the velocity with which the ball strikes the ground and the height of the tower in meters.
vf=vi+aΔt
=0-9.8*3
=-29.4 m/s
Distance
=average velocity*time
=-14.7*3
=-44.1 m
78
44.1
To find the velocity with which the ball strikes the ground and the height of the tower, we can use the equations of motion.
First, let's find the velocity of the steel ball just before it hits the ground. We can use the equation of motion:
v = u + at
Here,
v = final velocity (which is 0, as the ball comes to rest on hitting the ground)
u = initial velocity (which we need to find)
a = acceleration due to gravity (which is approximately 9.8 m/s^2)
t = time taken (which is 3.0 seconds)
Plugging in the values, we get:
0 = u + (9.8 m/s^2) * (3.0 s)
Rearranging the equation, we find:
u = - (9.8 m/s^2) * (3.0 s)
Calculating the value, we get:
u = -29.4 m/s
Thus, the initial velocity of the ball is -29.4 m/s (negative because it is directed downwards).
Next, to find the height of the tower, we can use the equation of motion:
s = ut + (1/2)at^2
Here,
s = height of the tower
u = initial velocity (-29.4 m/s)
t = time taken (3.0 seconds)
a = acceleration due to gravity (9.8 m/s^2)
Substituting the values and solving for s, we get:
s = (-29.4 m/s) * (3.0 s) + (1/2) * (9.8 m/s^2) * (3.0 s)^2
Calculating the value, we find:
s = -88.2 m + 44.1 m
s = -44.1 m
Since the height cannot be negative, we take the absolute value of s:
s = 44.1 m
Therefore, the height of the tower is 44.1 meters and the velocity with which the ball strikes the ground is -29.4 m/s (negative sign indicates direction towards the ground).