Explanation:

u=25m/s

at maximum height v=0

(1)....using 2as = v^2--u^2

2(-9.8)s=-625

s=625/19.6=31.87m

(2)....time to reach maximum height is v=u+at

25=9.8t

t=25/9.8=2.5sec

after 2.5 sec ,object will fall in downward direction so after 1.5 sec its position is

s=. 1/2 (9.8)(1.5)^2=11.025 from top

Verified answer

Answer:

The total time taken by the object to reach the ground again is around 2.55 seconds, and after 4 seconds, the object is approximately 21.6 meters above the ground.

Certainly! Here are the calculations:

1. **Total time taken by the object to reach the ground again:**

[tex] \[ t = \frac{v - u}{a} \] \[ t = \frac{0 - 25}{-9.8} \] .………\[ t ≈ 2.55 \, \text{seconds} \][/tex]

2. **Position of the object from the ground after 4 seconds:**

[tex] \[ s = ut + \frac{1}{2}at^2 \]

\[ s = (25 \times 4) + \frac{1}{2}(-9.8) \times (4)^2 \]...\[ s ≈ 21.6 \, \text{meters} \][/tex]