Water droplets are coming from an open tap at | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the time interval between two consecutive drops be $T$. At $t = 4 \, s$, the first drop has fallen for $4 \, s$. The distance covered by the f

Vedclass · Accepted Answer

$1 \, 	ext{drop/second}$

Vedclass · Answer

(A) Let the time interval between two consecutive drops be $T$. At $t = 4 \, s$, the first drop has fallen for $4 \, s$. The distance covered by the first drop is $h_1 = \frac{1}{2} g (4)^2 = \frac{1}{2} 	imes 9.8 	imes 16 = 78.4 \, m$. The second drop was released $T$ seconds after the first, so its time of fall is $(4 - T) \, s$. The distance covered by the second drop is $h_2 = \frac{1}{2} g (4 - T)^2$. The spacing between the two drops is given as $h_1 - h_2 = 34.3 \, m$. Substituting the values: $78.4 - \frac{1}{2} 	imes 9.8 	imes (4 - T)^2 = 34.3$. $78.4 - 4.9(4 - T)^2 = 34.3$. $4.9(4 - T)^2 = 44.1$. $(4 - T)^2 = 9$. $4 - T = 3 \Rightarrow T = 1 \, s$. Since the time interval between drops is $1 \, s$, the rate is $1 \, 	ext{drop/second}$.

Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at $4^{th} \, s$ after its fall and the next droplet is $34.3 \, m$. At what rate are the droplets coming from the tap? (Take $g = 9.8 \, m/s^2$)

Similar Questions

$A$ player throws a ball upwards with an initial speed of $29.4 \; m s^{-1}$. To what height does the ball rise and after how long (in seconds) does the ball return to the player's hands? (Take $g = 9.8 \; m s^{-2}$ and neglect air resistance).

$A$ very large number of balls are thrown vertically upwards in quick succession in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is $5\, m$,the number of balls thrown per minute is (take $g = 10\, ms^{-2}$):-

$A$ juggler maintains four balls in vertically upward motion. He throws the next ball every $1/4 \, s$. For the show to continue,what should be the height to which he throws each ball? (Assume $g = 10 \, m/s^2$)

$A$ body is projected vertically upward from the surface of the Earth. If the upward direction is taken as positive,then the acceleration of the body during its upward and downward journey are respectively:

If a stone thrown vertically upwards from a bridge with an initial velocity of $5 \,m \,s^{-1}$ strikes the water below the bridge in a time of $3 \,s$, then the height of the bridge above the water surface is (Acceleration due to gravity $= 10 \,m \,s^{-2}$) (in $\,m$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module