A-ball-is-thrown-vertically-upwards-with-a-velocity-of-10-ms-1-from-a-point-75m-above-the-ground-How-long-will-it-take-the-ball-to-strike-the-ground-

Question Number 153041 by nadovic last updated on 04/Sep/21

A ball is thrown vertically upwards

with a velocity of 10 {m s}^{- 1} from a point

75 m above the ground . How long will

it take the ball to strike the ground ?

Answered by mr W last updated on 04/Sep/21

- 75 = 10 t - \frac{10}{2} t^{2}

t^{2} - 2 t - 15 = 0

(t - 5) (t + 3) = 0

\Rightarrow t = 5 s e c o n d s

Commented by nadovic last updated on 05/Sep/21

Thank you Sir

Answered by physicstutes last updated on 04/Sep/21

Using the kinematic equation

y = y_{0} + v_{0} t + \frac{1}{2} {g t}^{2}

\Rightarrow y = 75 + 10 t - \frac{10}{2} t^{2} [assuming g = 10 m / s^{2}]

when it reaches the ground, y = 0

\Rightarrow 5 t^{2} - 10 t - 75 = 0

\Rightarrow t^{2} - 2 t - 15 = 0

\Rightarrow (t - 5) (t + 3) = 0

t = 5 seconds since t > 0

Commented by nadovic last updated on 05/Sep/21

Thank you Sir

Answered by talminator2856791 last updated on 04/Sep/21

\max height = 10 \times \frac{10}{9.8} \div 2 + 75

v \times \frac{v}{9.8} \div 2 = 10 \times \frac{10}{9.8} \div 2 + 75

(the point is to find maximum

velocity then t follows)

v^{2} = 1570

v = \sqrt{1570}

t = \frac{10}{9.8} + \frac{\sqrt{1570}}{9.8}

t = 5.06