An object whose mass is $m$ is crashing into the ground and leaping with the same velocity. If the duration of the interaction between the object and the ground is $0.1$ seconds, determine the force ground applied to the object.
I know that
$$\Delta P_y = F_y \times \Delta t$$
$$\Delta P_y = (N-mg) \times \Delta t$$
$$2mv_0 \sin \theta = (N-mg) \times \Delta t$$
Hence we get
$$N = 220N$$
According to answer key, it seems incorrect.
Usually this is an impulse force which is triangular in shape. However, assuming its a constant force (rectangular), then it should be:
The normal velocity is $5m/s$ to the ground. It reduces to $0 m/s$ in $.05$ seconds. So $F = ma$ therefore: $$ F = 2\cdot \frac{5}{.05} = 200 N$$