A ball is thrown upward from the ground with an initial speed of 25 m/s; at the same instant, a ball is dropped from rest from a building 15 m high. After how long will the balls be at the same height?
X1 = 25 t - 4.9 t^2
X2 = 15 - 4.9 t^2
Set X1 = X2 and solve for t.
25t = 15
t = 3/5 seconds
lolno
To find the time at which both balls will be at the same height, we need to determine the height of each ball as a function of time and then find the time when their heights are equal.
Let's start by finding the expression for the height of the ball thrown upward. We can use the formula for the height of an object in free fall:
h(t) = h0 + v0*t - (1/2)*g*t^2
Where:
h(t) is the height as a function of time
h0 is the initial height (which in this case is 0 since the ball is thrown upward from the ground)
v0 is the initial velocity (in this case 25 m/s)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time
For the ball thrown upward:
h1(t) = 0 + 25t - (1/2)*9.8*t^2
= 25t - 4.9*t^2
Now, let's find the expression for the height of the ball dropped from the building. Since the ball is dropped from rest, its initial velocity is 0. We can use the formula:
h(t) = h0 + v0*t - (1/2)*g*t^2
Where:
h(t) is the height as a function of time
h0 is the initial height (which in this case is 15 m since the ball is dropped from a 15 m high building)
v0 is the initial velocity (which is 0 for the ball dropped from rest)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time
For the ball dropped from the building:
h2(t) = 15 + 0*t - (1/2)*9.8*t^2
= 15 - 4.9*t^2
To find the time at which both balls are at the same height, we need to set the height functions h1(t) and h2(t) equal to each other:
25t - 4.9*t^2 = 15 - 4.9*t^2
Simplifying the equation:
25t = 15
Now, we can solve for t:
t = 15/25
t = 0.6 seconds
Therefore, the balls will be at the same height after 0.6 seconds.