MadeEasy Test Series 2019: General Aptitude - Speed Time Distance

Question

A ball thrown upward satisfy the equation s=160 t – 20 t^2, where S is in m (meter) and t is in sec. what is the maximum height achieved by the ball ?

(a) 80

(b) 160

(c) 240

(d) 320

According to me..

Here s=height reached by the ball.

t=time taken by the ball to reach the particular height.

A/c to equation of motion;

s=ut+1/2at^2 &

v=u+at

Here, a=acceleration due to gravity, i.e., g=9.8m/s^2

Since the acceleration is is the opposite direction, we have to use the equations of motions as-

s=ut-1/2gt^2 &

v=u-gt

Here v=final velocity

Since at maximum height the stone will have a velocity of zero (since it will stop at the highest point).

So, v=0

than 

can anyone tell me where I am wrong ?

Comments

bro did these motion formulas in 12th didnt remember that much ...............

but what u do just differentiate it and assign to 0 ( apply max. eq. so that it verify that it is max) u get i think 4 as t

then apply 4 we get 320

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okk thanx broo @Deepanshu

Answer 1

Here we can find it by simply differentiation.

Do as follows:

S= 160t - 20t^2

s'= 160- 40t= 0

t=4

now, s''= -40 , we will get maximum value here.

s= 160*4 - 20*16= 320

Answer 2

the answer is very simple

we know s=ut+1/2*a*t^2

from the equation given we can say u=160 and a=-40

we know h=-u^2/2a(from v^2-u^2=2gh v=0)

substituting we will get h as -320

so the ball will go 320m