SATHEE: Kinematics Question 414

Kinematics Question 414

Question: A body dropped from top of a tower fall through 40 m during the last two seconds of it’s fall. The height of tower it’s $(g = 10 m / s^{2})$

Options:

A) 60 m

B) 45 m

C) 80 m

D) 50 m

Show Answer

Answer:

Correct Answer: B

Solution:

Let the body fall through the height of tower in t seconds. From, $D_{n} = u + \frac{a}{2} (2 n - 1)$ we have, total distance travelled in last 2 second of fall it’s $D = D_{t} + D_{(t - 1)}$

$= [0 + \frac{g}{2} (2 t - 1)] + [0 + \frac{g}{2} [2 (t - 1)] - 1]$

$= \frac{g}{2} (2 t - 1) + \frac{g}{2} (2 t - 3) = \frac{g}{2} (4 t - 4)$

$= \frac{10}{2} \times 4 (t - 1)$

$or, 40=20 (t - 1) or t=2+1=3s$ distance travelled at t second is $s=ut+ \frac{1}{2} a t^{2} = 0 + \frac{1}{2} \times 10 \times 3^{2} = 45 m$

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Kinematics Question 414

Question: A body dropped from top of a tower fall through 40 m during the last two seconds of it’s fall. The height of tower it’s (g=10m/s2)

Options:

Answer:

Solution:

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Question: A body dropped from top of a tower fall through 40 m during the last two seconds of it’s fall. The height of tower it’s $(g = 10 m / s^{2})$