What are the solutions to the equation 0=0+25t-(0.5*9.8t^2) ?

The solutions to the equation 0=0+25t-(0.5*9.8t^2) are t=0,t=(50)/(9.8)

Find the zeros of 0=0+25t-(0.5*9.8t^2)

The zeros of 0=0+25t-(0.5*9.8t^2) are t=0,t=(50)/(9.8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0 = 0 + 25 t - (0.5 \cdot 9.8 t^{2})

t = 0, t = \frac{50}{9.8}

Decimal Notation

t = 0, t = 5.10204 \dots

Solution steps

0 = 0 + 25 t - (0.5 \cdot 9.8 t^{2})

Expand

0 + 25 t - (0.5 \cdot 9.8 t^{2}) : - 4.9 t^{2} + 25 t

0 = - 4.9 t^{2} + 25 t

Switch sides

- 4.9 t^{2} + 25 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 25 \pm 2 5 ^{2} - 4 ( - 4.9 ) \cdot 0}{2 ( - 4.9 )}

2 5^{2} - 4 (- 4.9) \cdot 0 = 25

t_{1, 2} = \frac{- 25 \pm 25}{2 ( - 4.9 )}

Separate the solutions

t_{1} = \frac{- 25 + 25}{2 ( - 4.9 )}, t_{2} = \frac{- 25 - 25}{2 ( - 4.9 )}

t = \frac{- 25 + 25}{2 ( - 4.9 )} : 0

t = \frac{- 25 - 25}{2 ( - 4.9 )} : \frac{50}{9.8}

The solutions to the quadratic equation are:

t = 0, t = \frac{50}{9.8}

lo g_{\frac{1}{5}} (5)

16 b^{4} + 8 b^{2} + 20 b

\frac{a}{4} - \frac{2 a - 2}{3} < \frac{2 a}{4}

1. 5^{2}

\frac{1}{4} (4)^{2}

What are the solutions to the equation 0=0+25t-(0.5*9.8t^2) ?
The solutions to the equation 0=0+25t-(0.5*9.8t^2) are t=0,t=(50)/(9.8)
Find the zeros of 0=0+25t-(0.5*9.8t^2)
The zeros of 0=0+25t-(0.5*9.8t^2) are t=0,t=(50)/(9.8)