Question

A object is thrown verticly upwards and rises to a height if 40m calculating (a) the velocity with which the object was thrown upwards (b) the time taken by the object to reach the height point given g=9.8ms-2​

Answer 1

Answer:

$the \: object \: was \: thrown \\ \: upwards \\ \: with \: 28ms {}^{ - 1} velocity \: and \: \: \\ took \\ \ \: 2.86seconds \: to \: reach \: the \: \\ highest \: point.$

Explanation:

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: given$

height (h) = 40m

$acceleration \: due \: to \: gravity \: (a) = ( - g) = - 9.8ms {}^{ - 2}$

$final \: velocity(v) = 0$

$(a)using \: 3 {rd} \: equation \: of \: motion \\ ...$

$v {}^{2} = u {}^{2} + 2gh$

$0 = u {}^{2} - 2 \times 9.8 \times 40$

$u {}^{2} = 19.6 \times 40$

$u = \sqrt[]{19.6 \times 40}$

$therefore \: \\ initial \: velociy(u) = 28ms {}^{ - 1}$

$(b) \: let \: the \: time \: taken \: by \: the \: \: \\ object \: to \: reach \: the \: highest \: point \\ \: be \: = t$

$using \: 1 {}{st} \: equation \: of motion...$

$v = u + gt$

$0 = 28 - 9.8t$

$9.8t = 28$

$t = \frac{28}{9.8} = 2.86 \: s$

Answer 2

Answer:

नमस्ते

Explanation:

hi. HE'LLO ......... good morning