Question

A ball is thrown with an upward velocity of 5 m/???? from the top of a 10 m high building. One second later. Another ball is thrown vertically from the ground with a velocity of 10 m/????. Determine the height from the ground where the two balls pass each other.

Answer 1

Answer:

$y=4.54\ m$

Explanation:

Vertical Throw

When an object is launched in free air (no friction) and its movement is confined to the vertical axis, the only force acting is the weight of the object. Let's take the positive direction of the movement upwards and negative downwards. When the object is thrown upwards, the initial speed is positive. When the object starts to fall to ground, the speed is negative.

Given an initial speed, the final speed at a time t is given by

$v_f=v_o-gt$

And the height of the object is

$\displaystyle y=y_o+v_ot-\frac{gt^2}{2}$

The first ball is thrown up from a height yo=10 m with vo=5 m/s. The height at any time t is

$y_1=10+5t-4.9t^2$

One second later, a second ball is thrown up from the ground (y0=0) with vo=10 m/s. Its height is

$\displaystyle y_2=0+10(t-1)-4.9(t-1)^2$

$y_2=10t-10-4.9(t-1)^2$

We must find the value of t where

$y_1=y_2$

Or, equivalently

$10t-10-4.9(t-1)^2=10+5t-4.9t^2$

Operating

$5t-20-4.9(t^2-2t+1)=-4.9t^2$

Simplifying

$5t-20-4.9t^2+9.8t-4.9=-4.9t^2$

$14.8t=24.9$

Solving for t

$t=24.9/14.8=1.6824 \ sec$

$t=1.6824\ s$

The height of the two balls can be computed with any of the formulas

$y_1=10+5t-4.9t^2$

$y_1=10+5(1.6824)-4.9(1.6824)^2$

$y_1=4.54\ m$

If we computed with the other formula:

$y_2=10(1.6824)-10-4.9(1.6824-1)^2$

$y_2=4.54\ m$

They result the same as expected

Answer 2

The two balls undergoing a vertically upward motion would pass each other at a height of $4.57\,m$ from the ground.

The answer is explained as follows.

Given that a ball is thrown upward with an initial velocity,

• $u = 5m/s$

Initially, the position of the ball is;

• $s_i = 10\,m$

• And let $s_f$ be the final position of the ball.

From the kinematics equation, we have;

• $s_f - s_i = ut + \frac{1}{2}at^2$

• But in the case of a vertically upward motion, the acceleration due to gravity acts downward, so here, $a=-g=-9.8m/s^2$.

Now, substituting all the values, we get;

$s_f - 10=5t-(\frac{1}{2} \times9.8)t^2$

• $\implies s_f = 10+5t-4.9t^2$

In the case of the second ball,

• $s_i = 0$, (as the ball is thrown from the ground).

And the initial velocity,

• $u=10\,m/s$

• The time here, is a second later than the first ball, i.e.; $(t-1)\,$ seconds.

Here, from the kinematics equation, we have;

• $s_f - s_i = ut + \frac{1}{2}at^2$

Substituting all the known values,

$s_f -0= 10(t-1) -( \frac{1}{2} \times 9.8)(t-1)^2$

$\implies s_f = 10t-10-4.9(t^2 - 2t+1)=10t-10-4.9t^2+9.8t-4.9$

• $s_f =-4.9t^2+19.8t-14.9$

• When the two balls pass each other, they have the same the same final position ( $s_f$ ).

Now, equating the final position of both the balls, we get;

$\implies 10+5t-4.9t^2 =-4.9t^2+19.8t-14.9$

$24.9=14.8t$

$\implies t=\frac{24.9}{14.8}=1.68\,s$

Now, re-substituting the value of $t$ in any of the equation for final position, we get;

$s_f = 10+(5\times1.68)- (4.9\times 1.68 \times 1.68)=4.57\,m$.

So, the balls will meet at a height of 4.57 meters from the ground.

Learn more about vertical motion here:

https://brainly.com/question/13665920