A rock is thrown vertically upward with a speed of 12.0m/s.Exactly 1.00s later a ball thrown up vertically along the same path with a speed of 18.0m/s.
a. At what time will the ball collide with the rock?
for the rock ... h = -4.9 t^2 + 12.0 t
for the ball ... h = -4.9 (t - 1)^2 + 18.0 (t - 1)
set the two heights equal , and solve for t
To find the time at which the ball collides with the rock, we need to determine when their paths intersect.
Let's start by analyzing the motion of the rock. The acceleration due to gravity acts in the downward direction, opposing the upward motion of the rock.
Using the equation of motion:
Final velocity (v) = Initial velocity (u) + Acceleration (a) * Time (t)
At the highest point, the velocity of the rock will be zero because its motion will change from upward to downward. Thus, the final velocity is zero (v = 0). The initial velocity is 12.0 m/s (u = 12.0 m/s), and since the rock is moving upwards, the acceleration due to gravity is -9.8 m/s² (a = -9.8 m/s²). We need to find the time it takes for the rock to reach this point, which we'll call t1.
0 = 12.0 m/s + (-9.8 m/s²) * t1
Solving this equation will give us the time taken by the rock to reach its highest point (t1).
Next, let's analyze the motion of the ball. It is thrown up with an initial velocity of 18.0 m/s. Since it's moving in the same path as the rock, its acceleration due to gravity is also -9.8 m/s².
We need to calculate the time it takes for the ball to reach the same height as the rock, which we'll call t2. The initial velocity of the ball is 18.0 m/s (u = 18.0 m/s), and the acceleration is -9.8 m/s² (a = -9.8 m/s²). The final velocity is not given, but we know that it will also be zero when the ball reaches its highest point.
0 = 18.0 m/s + (-9.8 m/s²) * t2
Solving this equation will give us the time taken by the ball to reach the same height as the rock (t2).
Now, we need to find the time at which the ball collides with the rock. Since the ball is thrown exactly 1.00 second after the rock, we can express the time taken by t2 as t1 + 1.00 second:
t2 = t1 + 1.00 s
By substituting t1 + 1.00 second for t2 in the equation 0 = 18.0 m/s + (-9.8 m/s²) * t2, we can solve for t1, which will give us the time at which the ball collides with the rock.