# Question #81787

Dec 19, 2016

Range =$\textsf{1019 \textcolor{w h i t e}{x} m}$

$\textsf{T = 14.4 \textcolor{w h i t e}{x} s}$

#### Explanation:

To get the time of flight T use:

$\textsf{v = u + a t}$

To get the time t to reach the maximum height, take vertical components of motion:

$\textsf{0 = v \sin \theta - \text{gt}}$

$\therefore$$\textsf{t = \frac{v \sin \theta}{g} = \frac{100 \sin 45}{9.81} = 7.208 \textcolor{w h i t e}{x} s}$

Since the motion is symmetrical we can say that the total time of flight T will be given by:

$\textsf{T = 2 t = 2 \times 7.208 = 14.41 \textcolor{w h i t e}{x} s}$

To get the range use the horizontal component of velocity which is constant.

$\textsf{s = v \cos \theta \times t}$

$\therefore$$\textsf{s = 100 \cos 45 \times 14.41 = 1019 \textcolor{w h i t e}{x} m}$