Question

A ball is thrown with a speed v=6.9m/s at an angle θ=22◦ with respect to the horizontal ground. At the highest point in the motion, the strength of gravity is somehow magically doubled. What is the total horizontal distance traveled by the ball by the time it returns to the height from which it was thrown? (Ignore air resistance.)

Answer 1

$x=x_1+x_2$

$x_1=v_0\cdot\cos\alpha\cdot t_1=v_0\cdot\cos\alpha\cdot\frac{v_0^2\sin(2\alpha)}{2g}=$

$=6.9\cdot\cos22°\cdot\frac{6.9^2\sin(2\cdot22°)}{2\cdot9.81}=1.68m$

$h=\frac{v_0^2\sin^2(\alpha)}{2g}=\frac{6.9^2\sin^2(22°)}{2\cdot 9.81}=0.34m$

$x_2=v_0\cdot\cos(\alpha)\cdot\sqrt{\frac{2h}{(2g)}}=6.9\cdot\cos(22°)\cdot\sqrt{\frac{0.34}{9.81}}=1.19m$

$x=x_1+x_2=1.68+1.19=2.87m$