Question #158394

A projectile is lunched with 30m/s at an angle 60 degree with the horizontal. Find the position of the projectile and the magnitude and direction of its velocity at t=2s


1
Expert's answer
2021-01-25T13:55:00-0500

An object moving in the two-dimensional parabolic path is a projectile motion. It moves in that trajectory because it is affected only by gravitational acceleration. Projectile Motion is described by two-dimensional kinematic equations.

Position

xt=xo+v0cos(θo)tx_t = x_o+v_0cos(\theta_o)t

yt=yo+vosin(θo)ty_t = y_o + v_osin(\theta_o)t

Velocity

vx=vocos(θo)v_x = v_ocos(\theta_o)

vy=vosin(θo)gtv_y = v_osin(\theta_o)-gt

The given in the problem:

v0=30ms    θ0=60o    g=9.8ms2v_0=30\frac{m}{s} \space \space \space \space \theta_0=60^o \space\space\space\space g = 9.8 \frac{m}{s^2}

vx=30cos60o=15ms     vy=15sin6009.82=6.6msv_x = 30*cos60^o = 15\frac{m}{s} \space\space\space\space\space v_y =15*sin60^0-9.8*2 = -6.6\frac{m}{s}

The negative answer indicates it is moving downward.

Using Pythagoras: v=vx2+vy2=152+(6.6)2=16.4ms2v =\sqrt{v_x^2+v_y^2} = \sqrt{15^2+(-6.6)^2} = 16.4\frac{m}{s^2}



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