Question

A ball is through vertically upwards with a velocity of 49 m/s calculate 1. The maximum height to which it rises. 2. The total time it takes to return to the surface of the earth

Answer 1

Solution :

$\rm{Given}\begin{cases}\sf{\bullet\: \: initial \: velocity \: of \: ball = 49m/s}\\\sf{\bullet\:acceleration \: due \: to \: gravity = 9.8m/ {s}^{2} }\\\sf{\bullet\:velocity \: at \: maximum \: height = 0} \end{cases}$

$\underline{ \large \purple{ \mathscr{\dag\:A \bf{ccourding} \: to \: \mathscr {Q} \bf{uestion} ....}}}$

we need to find the maximum height to which the ball rises :

By using 3rd equation of motion :

• v² = u² + 2as

=> 0² = 49² + 2 ×(- 9.8) × h

=> 0 = 2401 - 19.6h

=> - 2401/-19.6 = h

=> height = 122.5m

2)The total time it takes to return to the surface of the earth

By using 1st equation of motion :

• v = u + at

=> 0 = 49 - 9.8 × t

=> - 49/-9.8 = t

=> t = 5s

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