Question

An object is thrown upward from a tower crane relative to the ground. The said object reaches a high point of 5.10 m where you release it. If it hits the ground at 5.09 s where you release it, compute for its initial velocity going up and the height of the tower crane.

a.
Vi = 10 m/s ; d = 56 m


b.
no correct answer among the given choices


c.
Vi = 8.25 m/s ; d = 70 m


d.
Vi = 8.25 m/s ; d = 82 m


e.
Vi = 10 m/s ; d = 76 m

Answer 1

Answer:

answer is d

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Answer 2

The ball leaves form the top of the crane at a speed of 10 m/s and height of the crane is 76 m (e).

Given:

Height reached by the ball $= 5.1 m$

Time of flight $= 5.09 s$

To find:

Initial velocity $(u)$ of the object while going up   ; and

Height of the tower crane.

Solution:

Step 1

• We have been given that the object thrown up with some initial velocity (say u) reaches a point $x,5.1m$ above the crane whose height is say $AB=h$. We also know that the final velocity of the object at $x$ ill be 0 since at that point acceleration due to gravity and therefore the potential energy will be maximum.

Now, we have

$v=0m/s$   $;$  $s=h=5.1m$   $;$  $a=-g=10m/s^{2}$

Hence, the initial velocity of the ball will be

$v^{2}- u^{2}=2as$

$v^{2}- u^{2}=2gh$

Substituting the given values, we get

$-u^{2}=2(-10)(5.1)$

$u^{2}=102$

Hence, $u=10.1m/s$

Step 2

Now, time required by the object to reach the height $x$ will be

$s=ut+\frac{1}{2}at^{2}$

$5.1=(10.1)t+\frac{1}{2}(-10)t^{2}$

$5t^{2} -10.1t+5.1=0$

Solving this, we get, $t=1.02s$

Hence, the time required by the object to reach $x$ is $1.02 s$.

Step 3

We have been given that time required by the object to reach a height $x$ and then to the foot of crane at is $5.09 s$.

and we have calculated the time required for reaching from $h$ to $x$ hence, time for travelling distance $x$ will be  $(5.09-1.02)$ that is $4.07 s$.

Now, we have

height = $x$   $;$  $u=0m/s$    $;$  $a=g=10m/s^{2}$   $;$  $t=4.07s$

Therefore, value of height $x$ will be

$x=ut+\frac{1}{2}at^{2}$

Substituting the given values, we get

$x=(0)t+\frac{1}{2}(10)(4.07)^{2}$

$x=81.8m$

x is the total height reached by the object, therefore, the height $h$ of the crane will be

$h+5.1=81.8$

$h=76.7m$

Final answer:

Hence, the ball leaves from the top of the crane at a speed of 10 m/s and the height of the crane is 76.7 m.

The closest resembling option is (e).