## Engineering Mechanics: Statics & Dynamics (14th Edition)

$v=20.2ft/s$ $h=6.36ft$
We can determine the required speed and the distance as follows: We know that: $\Sigma F_y=0$ $\implies N_cos\theta=0$ $\implies N-(800cos 9.04)=0$ $\implies N=790.06lb$ First, we calculate the tangential velocity of the car $mr_1v_1+\int_0 ^t rNsin\theta=mrv_t$ $\implies 0+\int_0^4 (8\times 790.06sin 9.04)dt=(\frac{800}{32.2})8v_t$ $\implies v_t=19.99ft/s$ Now we determine the final speed of the car $v_2=\frac{v_t}{cos\theta}$ $\implies v_2=\frac{19.99}{cos 9.04}$ $\implies v_2=20.2ft/s$ According to the principle of work and energy $\frac{1}{2}mv_1^2+Wh=\frac{1}{2}mv_2^2$ We plug in the known values to obtain: $\frac{1}{2}(\frac{800}{32.2})(0)^2+800h=\frac{1}{2}(\frac{800}{32.2})(20.2)^2$ This simplifies to: $h=6.36ft$