Question

Ball-A is thrown vertically up from the ground at time t =0 with an initial velocity of 60 m/sec. Ball-B is thrown vertically up from the ground at time t = 0 seconds with an initial velocity of 45 m/sec. Calculate the maximum distance (D) between the two balls and the time (t) that this occurs at. Assume g = 10 m/sec^2.

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Answer 1

Ball A is thrown vertically up from the ground with an initial velocity of 60 m/s.

• Initial speed of ball A = 60 m/s

Ball B is thrown vertically up from the ground with an initial velocity of 45 m/s.

• Initial speed of ball B = 45 m/s

Both balls experience acceleration due to gravity g = 10 m/s² downwards.

• Acceleration of both balls = -10 m/s²

Now displacement of ball A at time t is,

$\small\text{ \longrightarrow\sf{s_A=60t+\dfrac{1}{2}(-10)t^2} }$

$\small\longrightarrow\sf{s_A=60t-5t^2}$

and displacement of ball B is,

$\small\text{ \longrightarrow\sf{s_B=45t+\dfrac{1}{2}(-10)t^2} }$

$\small\longrightarrow\sf{s_B=45t-5t^2}$

Now the distance between the two balls is,

$\small\longrightarrow\sf{d=s_A-s_B}$

$\small\longrightarrow\sf{d=(60t-5t^2)-(45t-5t^2)}$

$\small\longrightarrow\sf{d=15t}$

Here we see that the distance between the two balls increases with increase in t, till any of the two balls reach the ground as the time passes. So the maximum distance is obtained at the time when one ball reaches the ground.

We find the time at which each ball reaches ground, by equating their displacement zero, then identify which ball reaches ground first, i.e., whose time of reaching ground is the least. The maximum separation is obtained at that time.

For ball A,

$\small\longrightarrow\sf{s_A=0}$

$\small\longrightarrow\sf{60t-5t^2=0}$

$\small\longrightarrow\sf{t(60-5t)=0}$

Since t ≠ 0,

$\small\longrightarrow\sf{60-5t=0}$

$\small\longrightarrow\sf{t=12\ s}$

For ball B,

$\small\longrightarrow\sf{s_B=0}$

$\small\longrightarrow\sf{45t-5t^2=0}$

$\small\longrightarrow\sf{t(45-5t)=0}$

Since t ≠ 0,

$\small\longrightarrow\sf{45-5t=0}$

$\small\longrightarrow\sf{t=9\ s}$

So ball B reaches ground first. After 9 seconds ball B is at ground and ball A is arriving ground, then the separation decreases by time.

That's why maximum separation is at t = 9 seconds and is equal to,

$\small\longrightarrow\sf{D=15\times9}$

$\small\text{ \longrightarrow\underline{\underline{\sf{D=135\ m}}} }$

Answer 2

Answer:

Given :-

Ball-A is thrown vertically up from the ground at time t =0 with an initial velocity of 60 m/sec. Ball-B is thrown vertically up from the ground at time t = 0 seconds with an initial velocity of 45 m/sec.

To Find :-

Calculate the maximum distance (D) between the two balls and the time (t) that this occurs at. Assume g = 10 m/sec^2.

Solution :-

We know that

s = ut + ½ at²

s = (60)(t) + ½ × (-10)(t)²

s = 60t + (-5)t²

s = 60t - 5t²

case 2

s = (45)(t) + ½(-10)(t²)

s = 45t + (-5)t²

s = 45t - 5t²

on subtracting

60t - 5t²

45t - 5t²

______

15t

Now, we will find which ball lands first.

60t - 5t² = 0

5t² - 60t = 0

5t(t - 12) = 0

t - 12 = 0/5t

t - 12 = 0

t = 12 (i)

Similarly

45t - 5t² = 0

5t² - 45t = 0

5t(t - 9) = 0

t - 9 = 0/5t

t - 9 = 0

t = 9

Since t2 take less time than t1

Therefore

t = 9 sec

Distance = 15t

Distance = 15(9)

Distance = 135 m