A projectile is shot from the ground at a velocity of $4 \frac{m}{s}$ $4 m/s$ at an angle of $\frac{π}{6}$ $pi/6$ . How long will it take for the projectile to land?

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A projectile is shot from the ground at a velocity of $4 \frac{m}{s}$ $4 m/s$ at an angle of $\frac{π}{6}$ $pi/6$ . How long will it take for the projectile to land?

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Answer 1 · 2016-03-03T22:41:56

$4 \cdot \frac{\sqrt{3}}{5} m$ $4* sqrt(3)/5 m$

Explanation:

Let's start with Newton's equation $* F * = m \cdot * a *$ $**F** = m* **a**$ . As the problem is bi-dimensional, the previous equation can be split in the horizontal and vertical components

$F_{x} = m \cdot a_{x}$ $F_x = m*a_x$ for the horizontal component and

$F_{y} = m \cdot a_{y}$ $F_y = m*a_y$ for the vertical component.

The horizontal acceleration is null while the vertical acceleration is equal to gravity

$a_{x} = 0$ $a_x = 0$ ; $a_{y} = - g$ $a_y =-g$

and using the definition of acceleration

$a_{x} = \frac{d^{2} x}{{d t}^{2}}; a_{y} = \frac{d^{2} y}{{d t}^{2}}$ $a_x = (d^2x)/dt^2 ; a_y = (d^2y)/dt^2$

and integrating both part of equations

$x = v_{x 0} \cdot t; y = v_{y o} \cdot t - g \cdot \frac{t^{2}}{2} (1)$ $x= v_(x0)*t ; y= v_(yo)*t - g*t^2/2 (1)$

The projectile will land when $y = 0 \Rightarrow$ $y=0 =>$ introducing it in (1)

$0 = v_{y o} \cdot t_{l} - g \cdot \frac{t_{l}^{2}}{2}; t_{l} = 2 \cdot \frac{v_{y o}}{g}$ $0= v_(yo)*t_l - g*t_l^2/2 ; t_l = 2*v_(yo)/g$

and replacing in the equation or $x$ $x$

$x_{l} = 2 \cdot v_{x 0} \cdot \frac{v_{y 0}}{g}$ $x_l= 2*v_(x0)*v_(y0)/g$

The components of shot speed are

$v_{x 0} = v_{0} \cdot cos (\frac{π}{6})$ $v_(x0)= v_0 *cos(pi/6)$

$v_{y 0} = v_{0} \cdot sin (\frac{π}{6})$ $v_(y0)= v_0 *sin(pi/6)$

thus

$x_{l} = 2 \cdot \frac{v_{0}^{2}}{g} \cdot cos (\frac{π}{6}) \cdot sin (\frac{π}{6})$ $x_l= 2*v_0^2/g*cos(pi/6)*sin(pi/6)$

and replacing by the figures given in the problem

$x_{l} = 2 \cdot \frac{16}{10} m \cdot cos (\frac{π}{6}) \cdot sin (\frac{π}{6}) = \frac{16}{5} m \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} =$ $x_l= 2*16/10 m*cos(pi/6)*sin(pi/6) = 16/5m*1/2*sqrt(3)/2 =$ $= 4 \cdot \frac{\sqrt{3}}{5} m$ $=4* sqrt(3)/5 m$

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A projectile is shot from the ground at a velocity of $4 \frac{m}{s}$ $4 m/s$ at an angle of $\frac{π}{6}$ $pi/6$ . How long will it take for the projectile to land?

1 Answer

Explanation:

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