A ball is thrown straight up from the ground
with an initial velocity of 43.5 m/s; at the
same instant, a ball is dropped from the roof
of a building 15.4 m high.
After how long will the balls be at the
same height? The acceleration of gravity is
10 m/s^2
Answer in units of s.
the part about acceleration of gravity throws me off :(
just think of the equations of motion. You want the heights to be the same, so
0 + 43.5t - g/2 t^2 = 15.4 - g/2 t^2
The actual value of g does not matter, since both balls are affected by it. So, you wind up with
43.5t = 15.4
see the heights at
http://www.wolframalpha.com/input/?i=0+%2B+43.5t+-+5+t^2+%3D+15.4+-+5+t^2
The acceleration of gravity, denoted by the symbol "g," is a constant value that represents the rate at which an object falls under the influence of gravity. In this case, we are given that the acceleration of gravity is 10 m/s^2.
To find out when the balls will be at the same height, we need to determine the time it takes for each ball to reach that height.
Let's start with the ball that is thrown straight up. We can use the kinematic equation for displacement to calculate the time it takes to reach the maximum height. The equation is:
s = ut + (1/2)at^2
Where:
s = displacement (change in height)
u = initial velocity
t = time
a = acceleration (in this case, -g as the ball is moving upwards against gravity)
For the ball thrown straight up:
s = 0 (since it reaches maximum height and comes back down)
u = 43.5 m/s (initial velocity)
a = -10 m/s^2 (acceleration due to gravity)
Plugging in these values, the equation becomes:
0 = 43.5t + (1/2)(-10)t^2
Simplifying:
0 = 43.5t - 5t^2
Rearranging the equation:
5t^2 = 43.5t
Dividing both sides by t:
5t = 43.5
Solving for t:
t = 43.5/5 = 8.7 seconds
Therefore, it takes 8.7 seconds for the ball thrown straight up to reach its maximum height and come back down.
Now let's consider the ball that is dropped from the roof of the building. When an object is dropped, it falls freely under the influence of gravity. The time it takes for the ball to reach the ground can also be calculated using the same kinematic equation.
For the ball that is dropped:
s = 15.4 m (height of the building)
u = 0 (since it is initially at rest)
a = 10 m/s^2 (acceleration due to gravity)
Plugging in these values, the equation becomes:
15.4 = 0t + (1/2)(10)t^2
Simplifying:
15.4 = 5t^2
Dividing both sides by 5:
3.08 = t^2
Taking the square root of both sides:
t = √(3.08) ≈ 1.75 seconds
Therefore, it takes approximately 1.75 seconds for the ball dropped from the roof of the building to reach the ground.
To find when the balls will be at the same height, we need to add the time it takes for the ball thrown straight up to reach its maximum height (8.7 seconds) with the time it takes for the ball dropped from the roof to reach the ground (1.75 seconds):
8.7 seconds + 1.75 seconds = 10.45 seconds
Therefore, the balls will be at the same height after approximately 10.45 seconds.