velocity of the physics instructor with respect to bus
acceleration of the bus is given as
acceleration of instructor with respect to bus is given as
now the maximum distance that instructor will move with respect to bus is given as
so the position of the instructor with respect to door is exceed by
so it will be moved maximum by 3 m distance
Answer:
The maximum height covered is 3.25 m.
The horizontal distance covered is 9.81 m.
The total time in the air is 1.63 seconds.
Explanation:
The launch speed, .
Angle of launch with the horizontal,
So, the vertical component of the initial velocity,
.
The horizontal component of the initial velocity,
Let, t be the time of flight, to the horizontal distance covered
.
Not, applying the equation of motion in the vertical direction.
Where s is the displacement in time t, u is the initial velocity and a is the acceleration.
In this case, (from equation (i), s=0 (as the final height is same as the launch height) and
(negative sign is due to the downward direction).
seconds.
So, the total time in the air is 1.63 seconds.
From equation (i),
Total horizontal distance covered is
.
Now, for the maximum height, H, applying the equation of motion as
Here, v is the final velocity and v=0 (at the maximum height), and h=H.
So,
.
Hence, the maximum height covered is 3.25 m.