A body released from the top of a tower falls through half the height of the tower in 2 s. In what time shall the body fall through the height of the tower?

Question

A body released from the top of a tower falls through half the height of the tower in 2 s. In what time shall the body fall through the height of the tower? (A) 4 s (B) 3.26 s (C) 3.48 s (D) 2.828 s

Tardigrade Admin · Accepted Answer

u=0 (h/2)=(1/2) g(2)2 ⇒ h=4 g Now, time taken to fall through the height of the tower is t. h=(1/2) g t2 ⇒ 4 g=(1/2) g t2 or t2=8 ; t=√8=2.828 s

Q. A body released from the top of a tower falls through half the height of the tower in $2 s$ . In what time shall the body fall through the height of the tower?

4 s

3.26 s

3.48 s

2.828 s

$u = 0$
$\frac{h}{2} = \frac{1}{2} g (2)^{2} \Rightarrow h = 4 g$
Now, time taken to fall through the height of the tower is $t$ .
$h = \frac{1}{2} g t^{2} \Rightarrow 4 g = \frac{1}{2} g t^{2}$
or $t^{2} = 8; t = 8 = 2.828 s$

Q. A body released from the top of a tower falls through half the height of the tower in 2s. In what time shall the body fall through the height of the tower?

4 s

3.26 s

3.48 s

2.828 s

u=0 2h​=21​g(2)2⇒h=4g Now, time taken to fall through the height of the tower is t. h=21​gt2⇒4g=21​gt2 or t2=8;t=8​=2.828s

Q. A body released from the top of a tower falls through half the height of the tower in $2 s$ . In what time shall the body fall through the height of the tower?

$u = 0$
$\frac{h}{2} = \frac{1}{2} g (2)^{2} \Rightarrow h = 4 g$
Now, time taken to fall through the height of the tower is $t$ .
$h = \frac{1}{2} g t^{2} \Rightarrow 4 g = \frac{1}{2} g t^{2}$
or $t^{2} = 8; t = 8 = 2.828 s$