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Question: A boy wants to jump from building \(A\) to building \(B\). Height of the building \(A\) is \(25m\) a...

A boy wants to jump from building AA to building BB. Height of the building AA is 25m25m and that of building BB is 5m5m.Distance between buildings is 4m4m.Assume that the body jumps horizontally, then calculate minimum velocity with which he has to jump to land safely on building BB.

A. 6m/s6m/s
B. 8m/s8m/s
C. 4m/s4m/s
D. 2m/s2m/s

Explanation

Solution

In this question, first we will discuss about the second law of equation of motion and then find the time taken to jump and then use the horizontal distance to calculate the minimum velocity required to jump safely from building AA to building BB.

Complete step by step answer:
It is given that the boy wants to jump from building AA to building BB.The height of the building AA is 25m25m and the height of the building BB is 5m5m. The distance between the two buildings is 4m4m.The boy jumps horizontally.Let say the boy jumps at a velocity Vm/sVm/s.Hence the vertical velocity of the boy is 00 and the horizontal velocity is Vm/sVm/s as the boy jumps horizontally.
Now the vertical distance between the two buildings is 255=20m25 - 5 = 20m.
So we know the second law of equation of motion S=ut+12at2S = ut + \dfrac{1}{2}a{t^2} [where SS is the distance, uu is the initial velocity, tt is time and aa is acceleration due to gravity]
So the initial velocity of the boy is 00, a=10m/s2a = 10m/{s^2}, S=20mS = 20m the distance boy has to jump.
20=0×t+12×10t220 = 0 \times t + \dfrac{1}{2} \times 10{t^2}
After simplification we will get,
t=2st = 2s
The boy needs 2sec2\sec to jump from building AA to building BB.Now the horizontal distance is 4m4mand the horizontal velocity is Vm/sVm/s.
So, we can say that
Vt=4 V=4t V=42 V=2m/sVt = 4 \\\ \Rightarrow V = \dfrac{4}{t} \\\ \Rightarrow V= \dfrac{4}{2}\\\ \therefore V= 2m/s
So the minimum velocity to jump safely from AA to building BB is required 2m/s2m/s.

Hence option (D) is correct.

Note: As we know that acceleration is a vector quantity that has magnitude and direction. As the boy is falling down so the acceleration due to gravity is in the same direction of gravity so here acceleration is taken as positive.