Question

A biker is heading in the positive y direction with initial velocity 9.0 m/s. Her acceleration is given by a=[4.1,-2] m/s2.

How far is the biker from the point she started at the instance she reaches the maximum in y coordinate?

Answer 1

The maximum in y coordinate will be when

$$v_y=0=v-a_yt=9-2t\to t=4.5\ s$$

$$x=0.5a_xt^2=0.5(4.1)(4.5)^2=41.5125\ m$$

$$y=vt-0.5a_yt^2=(9)(4.5)-0.5(2)(4.5)^2\\=20.25\ m$$

$$s=\sqrt{41.5125^2+20.25^2}=46\ m$$