A balloon rises up with a constant net accele | vedclass

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

Question

(A) $1$. Initial motion of the balloon: Acceleration $a = 10\,m/s^2$,time $t_1 = 2\,s$. Velocity at $t_1 = 2\,s$ is $v = a t_1 = 10 	imes 2 = 20\,m/s

Vedclass · Accepted Answer

$(A \rightarrow r, B \rightarrow p, C \rightarrow s, D \rightarrow q)$

Vedclass · Answer

(A) $1$. Initial motion of the balloon: Acceleration $a = 10\,m/s^2$,time $t_1 = 2\,s$. Velocity at $t_1 = 2\,s$ is $v = a t_1 = 10 	imes 2 = 20\,m/s$. Height at $t_1 = 2\,s$ is $h_0 = \frac{1}{2} a t_1^2 = \frac{1}{2} 	imes 10 	imes 2^2 = 20\,m$.$2$. Motion of the particle after dropping: Initial velocity $u = 20\,m/s$,initial position $y_0 = 20\,m$,acceleration $a = -g = -10\,m/s^2$,time interval $\Delta t = 2\,s$.$3$. Height of particle from ground: $y = y_0 + u(\Delta t) + \frac{1}{2} a (\Delta t)^2 = 20 + 20(2) + \frac{1}{2}(-10)(2^2) = 20 + 40 - 20 = 40\,m$. Thus,$(A ightarrow r)$.$4$. Speed of particle: $v = u + a(\Delta t) = 20 + (-10)(2) = 0\,m/s$. Thus,$(B ightarrow p)$.$5$. Displacement of particle: $\Delta y = y - y_0 = 40 - 20 = 20\,m$. Wait,checking options: $C$ matches $s$ ($20$ $SI$ unit). Thus,$(C ightarrow s)$.$6$. Acceleration of particle: After release,only gravity acts,so $a = 10\,m/s^2$ (downward). Thus,$(D ightarrow q)$.$7$. Final match: $(A ightarrow r, B ightarrow p, C ightarrow s, D ightarrow q)$.

Column $I$	Column $II$
$(A)$ Height of particle from ground	$(p)$ $0$
$(B)$ Speed of particle	$(q)$ $10$ $SI$ unit
$(C)$ Displacement of particle	$(r)$ $40$ $SI$ unit
$(D)$ Acceleration of particle	$(s)$ $20$ $SI$ unit

Similar Questions

$A$ ball is thrown vertically upwards from the top of a tower at $4.9 \ m/s$. It strikes the pond near the base of the tower after $3 \ s$. The height of the tower is........$m$

$A$ ball is thrown vertically upwards with a velocity of $19.6 \, m/s$ from the top of a tower. The ball strikes the ground after $6 \, s$. The height from the ground up to which the ball can rise is given by $\left(\frac{k}{5}\right) \, m$. The value of $k$ is ..... (use $g = 9.8 \, m/s^2$)

$A$ ball of mass $0.5 \; kg$ is dropped from a height of $10 \; m$. The height from the ground,at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity,is $\dots \; m$. (Use $g = 10 \; m/s^2$).

$A$ ball is thrown upward with speed $10 \, m/s$ from the top of a tower and reaches the ground with a speed $20 \, m/s$. The height of the tower is ......... $m$. (Take $g = 10 \, m/s^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module