Question

please answer with full formula​

Answer 1

★ Question

A stone is thrown vertically upward with a speed of 40 m/s . the time interval for which particles was above 40 m from the ground

★ Answer :

u =40 m/s

S = 40 m

a = -g

time ,t = ?

$s = ut + \frac{1}{2} gt {}^{2}$

$40 = 40t - \frac{1}{2} \times 10 \times t {}^{2}$

$40 = 40t - 5t {}^{2}$

$5t {}^{2} - 40t + 40 = 0$

$t {}^{2} - 8t + 8 = 0$

after solving you get

$t = \frac{8 + \sqrt{32} }{2}$

$t = 4 + 2 \sqrt{2}$

t = 2.8+ 4

t = 6.8 sec

Answer 2

Question:

A stone is thrown vertically upward with a speed of 40 m/s . the time interval for which particles was above 40 m from the ground.

Answer:

u = 40 m/s

s = 40 m

a = - g

Time (t) = ?

s = ut + 1/2 gt²

40 = 40t - 1/2 × 10 × t²

40 = 40t - 5t²

t² - 8t + 8 = 0

t = 8 + √32/ 2

t = 4 + 2√2

t = 2.8 + 4

t = 6.8 sec.