Question

An object is thrown upward with an initial velocity 5m/s. Find the maximum height attained and time taken to reach the height. a=-9.8 m/s²​

Answer 1

ANSWER

Let u be the initial velocity and h be the maximum height attained by the stone.

v

1

2

​

=u

2

−2gh,

(10)

2

=u

2

−2×10×

2

h

​

100=u

2

−10h ....(i)

Again at height h,

v

2

2

​

=u

2

−2gh

(0)

2

=u

2

−2×10×h

u

2

=20h ... (ii)

So, from Eqs. (i) and (ii) we have

100=10h

h=10m

Answer 2

Explanation:

Given that :

Initial velocity (u) =5 m/s

Acceleration (a) = - 9.8 m/s^2

We have to find :

Maximum Height (H) = ?

Time for Maximum (t) = ?

Solution :

Final velocity should be zero for maximum height

V^2 =u^2 - 2gH

0 = 5^2 - 2 ×9.8 ×H

0 = 25 - 19.6 ×H

H = 25/19.6

H = 1.27 m Ans.

V = u + at

0 = 5 - 9.8 ×t

5 = 9.8 × t

t = 5/9.8

t =0. 5 se ans.