Question

At a certain instant, a ball is thrown downward with a velocity of 8.0 m/s from a height of 40 m. At the same instant, another ball is thrown upward from ground level directly in line with the first ball with a velocity of 12 m/s. Find (a) the time when the balls collide and (b) the height at which they collide. Take g = 10 m/s2

Answer 1

Answer:

(a) The two balls collide $2\; \rm s$ after launch.

(b) The height of the collision is $4\; \rm m$.

(Assuming that air resistance is negligible.)

Explanation:

Let vector quantities (displacements, velocities, acceleration, etc.) that point upward be positive. Conversely, let vector quantities that point downward be negative.

The gravitational acceleration of the earth points dowards (towards the ground.) Therefore, the sign of $g$ should be negative. The question states that the magnitude of $g\!$ is $10\; \rm m \cdot s^{-2}$. Hence, the signed value of $\! g$ should be $\left(-10\; \rm m \cdot s^{-2}\right)$.

Similarly, the initial velocity of the ball thrown downwards should also be negative: $\left(-8.0\; \rm m \cdot s^{-1}\right)$.

On the other hand, the initial velocity of the ball thrown upwards should be positive: $\left(12\; \rm m \cdot s^{-1}\right)$.

Let $v_0$ and $h_0$ denote the initial velocity and height of one such ball. The following SUVAT equation gives the height of that ball at time $t$:

$\displaystyle h(t) = \frac{1}{2}\, g \cdot {t}^{2} + v_0 \cdot t + h_0$.

For both balls, $g = \left(-10\; \rm m \cdot s^{-2}\right)$.

For the ball thrown downwards:

• Initial velocity: $v_0 = \left(-8.0\; \rm m \cdot s^{-1}\right)$.
• Initial height: $h_0 = 40\; \rm m$.

$\displaystyle h(t) = -5\, t^{2} + (-8.0)\, t + 40$ (where $h$ is in meters and $t$ is in seconds.)

Similarly, for the ball thrown upwards:

• Initial velocity: $v_0 = \left(12\; \rm m \cdot s^{-1}\right)$.
• Initial height: $h_0 = 0\; \rm m$.

$\displaystyle h(t) = -5\, t^{2} + 12\, t$ (where $h$ is in meters and $t$ is in seconds.)

Equate the two expressions and solve for $t$:

$-5\, t^{2} + (-8.0)\, t + 40 = -5\, t^{2} + 12\, t$.

$t = 2$.

Therefore, the collision takes place $2\, \rm s$ after launch.

Substitute $t = 2$ into either of the two original expressions to find the height of collision:

$h = -5\times 2^{2} + 12 \times 2 = 4\; \rm m$.

In other words, the two balls collide when their height was $4\; \rm m$.

Answer 2

The time the two balls collide is 0.4 seconds while the height at which they collide is 4m

The given parameters are :

Initial Velocity U = 8m/s

Height H = 40m

For the second ball, the initial velocity = 12m/s

a.) For the first ball, the height attained at the point of collision will be

h = ut + 1/2gt^2

h = 8t + 1/2 x 10t^2 ........ (1)

For the second ball, the height attained at the point of collision will be

h = 12t - 1/2 x 10t^2 .........(2)

Since the height will be the same for the two balls, equate the two equations

8t + 10t^2 = 12t - 10t^2

Collect the like term

8t - 12t = -5t^2 - 5t^2

-4t = -10^2

10t = 4

t = 4/10

t = 0.4s

b.) Substitute time t in any of the equation to find the height

h = 12(0.4) - 0.5 x 10(0.4)^2

h = 4.8 - 0.8

h = 4m

Therefore, the time the two balls collide is 0.4 seconds while the height at which they collide is 4m

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