Question
Question: A man runs across the roof-top of a tall building and jumps horizontally to land on the roof of the ...
A man runs across the roof-top of a tall building and jumps horizontally to land on the roof of the next building which is at a lower height than the first. If his speed is9ms−1, the horizontal distance between the two buildings is 10m and height difference is9m, will he be able to land on the next building? (g=10ms−2)
Solution
We are first going to draw a figure for the information given in the question, then, taking up the horizontal and the vertical motions for the man, we will calculate the time of flight for vertical motion, and then, the horizontal distance covered, which tells us whether he will be able to land or not.
Formula used: The equation of motion is given by
y=ut+21at2
Where, yis the distance covered, uis initially speed and tis time taken.
Complete step by step solution:
Let us first construct a figure for the given situation
The man’s speed is given as 9ms−1
The horizontal distance between the buildings is10mand the height difference is9m
Let the time taken for the man to fall from points Oto Abe tseconds.
Taking the vertical downward motion from OtoA, we get
{y_0} = 0 \\\
y = 9m \\\
{u_y} = 0 \\\
{a_y} = 10m{s^{ - 2}} \\\
We need to find time, t
The equation of motion is
y0=y=uyt+21ayt2
Putting the values in this, we get
−9=0×t+21×(−10)×t2
Finding the value oft, we get
t=109×2=1.8s
Now, considering the horizontal direction of motion,
vx=9ms−1
If Ris the horizontal distance travelled,
R=vxt
Putting the values in this,
R = 9 \times \sqrt {1.8} \\\ \Rightarrow R = 12m \\\
Thus, the man will be able to land on the building.
Note: It is important to note that the charged particles can be created or destroyed as long as the net charge on the particles before and after the process remains the same. But in that case also, charge is not being created but the charged particles only transfer charge from one system to another.