A stone is thrown vertically upwards from a cliff 20m high. After a time of 3 s it passes the edge of the cliff on its way down. Calculate a) the speed of projection b) the speed when it hits the ground c) the times when it is 10m above the top of the cliff d) the time it is 15m above the ground (take g=10ms^(−2) ).


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 54767 by pieroo last updated on 10/Feb/19

$A stone is thrown vertically upwards$ $from a cliff 20 m high . After a time of 3 s$ $it passes the edge of the cliff on its way$ $down . Calculate$ $a) the speed of projection$ $b) the speed when it hits the ground$ $c) the times when it is 10 m above the$ $top of the cliff$ $d) the time it is 15 m above the ground$ $(take g = {10 ms}^{- 2}) .$
Commented by pieroo last updated on 10/Feb/19

$NEED SOME HELP$
	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Feb/19

	$a) i n 3 s e c o n d s t h e s t o n e r e a c h (m a x i m u m h e i g h t a n d r e t u r n$ $t o t h e p o i n t o f p r o j e c t i o n) s o d i s p l a c e m e n t = 0$ $0 = u t - \frac{1}{2} {g t}^{2} s o u = \frac{g t}{2} = \frac{10 \times 3}{2} = 15 m / s e c$ $b) w h e n i t r e a c h a t t h e e d g e o f c l i f f i t a c q u i r e$ $v e l o c i t y = 0 + 10 \times \frac{3}{2} = 15 (v = u + a t t = \frac{3}{2} s e c)$ $r e q u i r e v e l o c i t y w h e n h i t g r o u n d$ $v_{g}^{2} = 15^{2} + 2 \times 10 \times 20 = 625$ $v_{g} = 25 m / s e c$ $c) 10 = 15 \times t - \frac{1}{2} \times 10 t^{2} [h = u t - \frac{1}{2} {g t}^{2}]$ $2 = 3 t - t^{2}$ $t^{2} - 3 t + 2 = 0 \to (t - 1) (t - 2) = 0$ $t = 1 s e c a n d 2 s e c$ $1 s e c w h i l e m o v i n g u p a n d 2 s e c w h e n i t r e a c h$ $s a m e p l a c e m o v i n g d o w n . . .$ $d) w h e n i t i s 15 m a b o v e g r o u n d t h a t m e a n s 5 m e t e r$ $b e l o w t h e b a s e o f c l i f f . . .$ $- 5 = 15 t - \frac{1}{2} \times 10 \times t^{2}$ $- 1 = 3 t - t^{2}$ $t^{2} - 3 t + 1 = 0$ $t = \frac{3 + \sqrt{9 - 4}}{2} = \frac{3 + \sqrt{5}}{2} s e c$ $p l s c h e k a n s w e r$ $i f n o t c o r r e c t p l s c o m m e n t . . .$
	Commented by pieroo last updated on 11/Feb/19

	$it is correct . thanks so much$
	Commented by tanmay.chaudhury50@gmail.com last updated on 11/Feb/19

	$m o s t w e l c o m e . . . m o r a l b o o s t . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum