Question

A projectile is launched with an initial speed of 39 m/s at an angle of 59° above the horizontal. What are the (a) magnitude and (b) angle of its velocity 2.0 s after launch (use the minus sign if the angle is below the horizontal)? What are the (c) magnitude and (d) angle of its velocity 5.0 s after launch (use the minus sign if the angle is below the horizontal)?

Answer 1

Answer:

$t=2$

$|v|=24.38693565m/s\\\alpha =34.54737228 ^{\circ}$

$t=5$

$|v|=25.41469214m/s\\\alpha =-37.78183103 ^{\circ}$

Explanation:

Let's use projectile motion equations, but first lets find the components of the initial speed vector:

$v_ox=v_o*cos(\alpha _o)=39*cos(59)=20.08648492m/s$

$v_oy=v_o*sin(\alpha _o)=39*sin(59)=33.42952473m/s$

Now let's find the magnitude and the angle of the velocity at t=2s, asumming g=9.8.

$v_x=v_ox=20.08648492m/s$

$v_y=v_oy-gt=33.42952473-(9.8*2)=13.82952473m/s$

Now the magnitude is given by:

$|v|=\sqrt{(v_x)^{2} +(v_y)^{2}} =24.38693565m/s$

and its angle:

$\alpha =arctan(\frac{v_y}{v_x} )=34.54737228 ^{\circ}$

Similarlyfor t=5

$v_x=v_ox=20.08648492m/s$

$v_y=v_oy-gt=33.42952473-(9.8*5)=-15.57047527m/s$

$|v|=\sqrt{(v_x)^{2} +(v_y)^{2}} =25.41469214m/s$

$\alpha =arctan(\frac{v_y}{v_x} )=-37.78183103 ^{\circ}$