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 and Answers Forum

All Questions Topic List
Mechanics Questions
Previous in All Question Next in All Question
Previous in Mechanics Next in Mechanics
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

	$\max height = 10 \times \frac{10}{9.8} \div 2 + 75$ $v \times \frac{v}{9.8} \div 2 = 10 \times \frac{10}{9.8} \div 2 + 75$ $(the point is to find maximum$ $velocity then t follows)$ $v^{2} = 1570$ $v = \sqrt{1570}$ $t = \frac{10}{9.8} + \frac{\sqrt{1570}}{9.8}$ $t = 5.06$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum