Question

a body is projected from a tower of height H which reaches the maximum height H as shown in Figure. the distance and displacement of the body are
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Answer 1

Answer:

The distance traveled by the body is $40$ meters, and the displacement is $42.43$ meters (upwards).

Explanation:

Assuming that the projectile is launched vertically upwards and air resistance is negligible, we can use the following kinematic equations to determine the distance and displacement of the body:

For the upward motion:

$v = u - gt\\H = ut - (1/2)gt^2$

For the downward motion:

$v^2 = u^2 - 2g(H - y)\\y = H - (1/2)gt^2$

where,

u = initial velocity (in the upward direction)

v = final velocity (in the downward direction)

g = acceleration due to gravity $(9.8 m/s^2)$

t = time taken to reach maximum height (H)

y = vertical displacement from the starting point

From the given figure, we can see that the initial velocity is $u = 0$, since the projectile is launched vertically upwards. The time taken to reach the maximum height H can be found using the equation:

$H = ut - (1/2)gt^2\\t = 2H / g$

Substituting the given values, we get:

$t = 2H / g = 2 \times 20 / 9.8 = 4.08$ seconds (approx.)

At the maximum height, the final velocity is v $= 0$, since the projectile momentarily comes to rest before falling back down. Therefore, the displacement from the starting point (which is also the final position) is equal to the vertical distance traveled during the upward journey, which is:

$y = H - (1/2)gt^2 = 20 - (1/2)(9.8)(4.08)^2 = 42.43$ meters (approx.)

The distance traveled by the projectile is equal to the total path length covered, which is the sum of the upward and downward journey distances:

distance $= 2H = 2 \times 20 = 40$ meters

Therefore, the distance traveled by the body is $40$ meters, and the displacement is $42.43$ meters (upwards).

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