Question

The velocity-time graph of a stone thrown
vertically upward with an initial velocity
30ms-1 is shown in the figure. The velocity
in the upward direction is taken as positive
and that in the downward direction as
negative. What is the maximum height to
which the stone rises?​

Answer 1

Answer: it is 45m

Explanation:

V2=U2+2as

V=0

U=30ms-1

a=-g

Therefore....

U2=2gs

30×30/2×10=s

S is 45

Answer 2

Answer:

45.87 $ms^{-2}$

Explanation:

Lets see the data we have

initial velocity = 30ms-1

acceleration = -9.81 ms-1

we need to calculate maximum height and at maximum height velocity will be zero

so the best formula to use will be

$v^{2} - u^{2}$ = 2as

v = 0

so      s = $\frac{u^{2} }{2a}$

          s = $\frac{30^{2} }{2 * 9.81}$

          s = 45.87 $ms^{-2}$