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This is one of our many quizzes on Free fall. Click on this to access more such competitive quizzes on Gravitation.

Free fall Quiz 2

physics

What is the height of the tower from which a ball takes $12$ seconds to free fall with an initial velocity is $1.2\,m/s$? $\text{(Take }g=9.8\,m/{{s}^{2}})$ 

 To calculate height of the tower from which a ball free falls, we use the equation of motion, suitable for the values given in the question i.e. the distance equation of motion and we get the height from the ball falls. 

 To calculate the height of the tower from which a ball free falls, we use the equation of motion, $ \\ H=ut+\dfrac{1}{2}g{{t}^{2}}  ----(i) \\ $ According to the question, $ \\ $ $u=1.2m/s$, $t=12$ seconds and $g=9.8m/{{s}^{2}} \\ $ Putting the above values in equation $(i)$, we get, $ \\ H=1.2\times 12+\dfrac{1}{2}\times 9.8\times {{(12)}^{2}} \\ \therefore H =720\,m$      

Consider a rock dropped from a height of $15$ m. If the initial speed of the rock is zero and it takes $2$ seconds to reach the ground, then at what speed does the rock land? $\text{(Take }g=9.8\,m/{{s}^{2}})$ 

 To find the speed of the rock when it lands, just before it touches the ground, we use the equation of motion $v=u+gt$. Since, the initial velocity and time taken is given in the question, we substitute the values in the above equation to find the final velocity before it lands. 

 To calculate the speed of the rock when it lands, we use the equation of motion, $ \\ v=u+gt ---(i) \\ $ According to the question, $ \\ $ $u=0m/s$, $t=2$ seconds and $g=9.8m/{{s}^{2}} \\ $ Putting the above values in equation $(i)$, we get, $ \\ v=0+9.8\times 2 \\ \therefore v =19.6\,m/s$

A kid throws a toy vertically towards the sky. If it takes 4.2 seconds to come back to its starting position, then what is the initial velocity of the ball? $\text{(Take }g=9.8\,m/{{s}^{2}})$ 

To calculate initial velocity of the ball thrown upwards, we use the equation of motion $S=ut+\dfrac{1}{2}g{{t}^{2}}$. Also, the ball comes back to its starting position, so the distance will be zero, put the values of time and g in the above equation to find the initial velocity. 

To calculate the initial velocity of the ball thrown upwards, we use the equation of motion, $ \\ S=ut+\dfrac{1}{2}g{{t}^{2}} ---(i) \\ $ According to the question, $ \\ $ $t=4.2$ seconds and $g=9.8m/{{s}^{2}}$ $ \\$ Putting the above values in equation (i), since, the ball comes back to its starting position then the distance will be zero, $ \\ \Rightarrow S=ut+\dfrac{1}{2}g{{t}^{2}} \\ \Rightarrow 0=u\times 4.2+\dfrac{1}{2}\times 9.8\times {{(4.2)}^{2}} \\ \therefore u=20.6\,m/s$  

 A small rocket is launched at an initial speed of $67\,m/s$ and it free falls to the ground to hit the starting point in $35$ seconds, then what is the maximum height it could reach? $\text{(Take }g=9.8\,m/{{s}^{2}})$ 

 To calculate the maximum height the rocket, we use the equation of motion, suitable for the values given in the question ${{v}^{2}}={{u}^{2}}+2gS$. And, we know that at maximum height the velocity of the rocket becomes zero i.e. final velocity is zero, hence, put the values in the above equation and solve for $S$. 

To calculate the maximum height of the rocket, we use the equation of motion, $ \\ {{v}^{2}}={{u}^{2}}+2gS  ---(i) \\ $ According to the question, $ \\ $ $u=67\,m/s$, $t=35$ seconds and $g=9.8\,m/{{s}^{2}}$ $ \\ $ Putting the above values in equation $(i)$ and at maximum height the velocity of the rocket becomes zero i.e. final velocity is zero, we get, $\Rightarrow {{v}^{2}}={{u}^{2}}+2gS \\ \Rightarrow 0={{(67)}^{2}}+2\times 9.8\times S \\ \therefore S=229\,m$       

A bomb will explode if it hits a target at the speed of $90\,m/s$. If this bomb is made to free fall from a height of $336$ m at an initial velocity of $2\,m/s$, what will happen to the bomb? $\text{(Take }g=9.8\,m/{{s}^{2}})$ 

The bomb will explode if it hits a target at the speed of $90\,m/s$, so if the final velocity of the bomb just before it hits the round is less that $90\,m/s$, then it will not explode. So, we have to calculate the final velocity of the bomb and compare the result with the given condition. 

The bomb will explode if it hits a target at the speed of $90\,m/s$, so if the final velocity of the bomb just before it hits the ground is less that $90\,m/s$, then it will not explode. So, we have to calculate the final velocity of the bomb, we use the equations of motion, $ \\ H=ut+\dfrac{1}{2}g{{t}^{2}} ---(i) \\ v=u+gt ----(ii) \\ $ According to the question, $ \\ $ $u=2m/s$, $H=336\,m$ and $g=10\,m/{{s}^{2}}$ $ \\ $ Now, to find the time taken to reach the round, we use equation $(i)$ and solve for $t$, we get, $S=ut+\dfrac{1}{2}g{{t}^{2}} \\ \Rightarrow 336=2t+\dfrac{1}{2}\times 10\times {{t}^{2}} \\ \Rightarrow t=8$ seconds $ \\ $ $[$We ignore the negative root of time, since time cannot be negative$]$ $ \\ $   Then, to find the final velocity, we use equation $(ii)$, $ \\ \therefore v=2+10\times 8=82\,m/s \\ $ So, we see that the final velocity is less than $90\,m/s$, at which the bomb explodes. Therefore, the bomb does not explode in this case.  