Question

a ball thrown up vertically returns to the thrower after 8 seconds. calculate velocity with which it was thrown up and maximum height and its position after 5s​

Answer 1

Answer:

${\red{Time \:taken \:to \:reach \:Maximum \:height :}}$ $\sf\dfrac{8}{2}} \sf{= 4s}$

v = 0 (at the maximum height )

a = - 9.8 m s-²

◆ a) Using, v = u + at, we get

0 = u +9.8 × 4

or, u = -39.2ms-¹

Thus, the velocity with which it was thrown up = -39.2ms-¹

◆ b) Using, 2aS = v² - u², we get

$\dfrac{{v}^{2}{-u}^{2}}{2a}$

$\sf\dfrac{{=0}\times{39.2}^{2}}{{2}\times{9.8}{ms}^{2}}$

= 78.4m

Maximum height = 78.4m

◆ c) Here, t = 5s. In 3 s, the ball reaches the maximum height and in 2 s it falls from the top.

Distance covered in 2 s from maximum height,

${S=}{{ut+\dfrac{1}{2}}{at}^{2}$

= $\sf{{0+}\dfrac{1}{2}}\times{9.8}\times{2}^{2}$

=  19.6m

Therefore, The ball will be 19.6 m below the top of the tower after 5 s.