Another Kinematics problem

**JonathanEyoon** · January 19th 2009, 07:48 AM

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.60

later.

What is the rocket's acceleration?

I'm wanting to use the 3 kinematics equations but I feel like i'm missing information. I know what the answer is already but I want to understand how to get it. It seems I can't simply use the equations for 1D kinematics so if anyone can show me i'd be much obliged.

**sumit2009** · January 19th 2009, 08:07 AM

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.60 later.

What is the rocket's acceleration?

Let accln. of the rocket=a

velocity of Bolt= velocity of rocket after 4 sec.= 4a .............upwards.................first eq of kinematics...

dist. covered by rocket in 4 sec.= 1/2 XaX 4X4= 8a ................................second eq of kinematics

now the Bolt falls from this height(8a) with accln. due to gravity in 6.6sec.

second eq of kinematics

s=ut-1/2 g t^2 .............................free fall
8a= 4a-1/2 X (9.8 X6.6X6.6)
solve this......

**skeeter** · January 19th 2009, 08:09 AM

while the bolt rides with the rocket (assuming the rocket started from rest) ...

$v = 4a$

$\Delta y = \frac{(4a)^2 - 0^2}{2a} = 8a$

when the bolt gets free, $a = -g$ ...

$\Delta y = 4a \cdot t - \frac{1}{2}gt^2$

$-8a = 4a(6.6) - \frac{1}{2}g(6.6)^2$

$-34.4a = - \frac{1}{2}g(6.6)^2$

$a = \frac{g(6.6)^2}{2(34.4)} \approx 6.2 \, m/s^2$

**Mush** · January 19th 2009, 08:10 AM

Originally Posted by JonathanEyoon

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.60

later.

What is the rocket's acceleration?

I'm wanting to use the 3 kinematics equations but I feel like i'm missing information. I know what the answer is already but I want to understand how to get it. It seems I can't simply use the equations for 1D kinematics so if anyone can show me i'd be much obliged.

Split the problem into two parts! Part 1 and part 2!

In part 1, the bolt is being accelerated from rest for 4 seconds.

$v_1 = u_1 + a_1t_1$

$v_1 = a_1t_1$

$v_1 = 4a_1$

We don't know what value of the acceleration, that's what we're trying to find! But we do know one thing about it, it's negative, since the rocket flies upwards.

Now let's have a look at part 2. In part 2 the bolt has a negative velocity $v_1 = 4a_1$ , and it accelerates towards the ground with $a_2 = 9.81 ms^{-2}$ . And it then hits the ground and has velocity 0, and this process takes 6.6 seconds!

$v_2 = u_2 + a_2t_2$

$0 = 4a_1 + 9.81(6.6)$

How's that?

**JonathanEyoon** · January 19th 2009, 08:47 AM

you guys are the best. I appreciate it!!!

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