Final answer:
The time of flight for a projectile with an initial velocity of 8i + 10j m/s, when projected on level ground and with gravity at 10 m/s², is 4 seconds.
Step-by-step explanation:
The time of flight of a projectile is completely determined by its vertical motion. For a projectile projected with initial velocity 8i + 10j m/s, the initial vertical component (10j m/s) and the acceleration due to gravity (g = 10 m/s2) are the important factors in determining the time it remains airborne. To calculate the time of flight, we use the kinematic equation for vertical motion, which is 0 = v_y - g*t, where v_y is the initial vertical speed and t is the time of flight.
Given that the initial vertical velocity (v_y) is 10 m/s and g is 10 m/s2, the time of flight is given by t = 2*v_y/g. Plugging in the values gives t = 2*10/10, which simplifies to t = 2 seconds. This represents the time for the projectile to reach the peak of its path. Since the projectile must come back down, the total time of flight is double that value, so t = 4 seconds. Therefore, the projectile spends 4 seconds in the air before landing back on the flat surface from which it was launched.