Given: T = 10s, Distance_y = 100m, Distance_x = 500m
If air resistance or friction is negligible
1.With what initial velocity was the cannonball launched?
Distance_x = V_x * T => 500 = V_x * T ==> V_x = 50 m / s
Distance_y = V_y * T - 0.5 * 9.8 * T^2 => 100 = V_y * 10 - 4.9 * 100 => 100 = V_y * 10 - 490 => 590 = V_y * 10 ==> V_y = 59 m / s
Initial Velocity = SQRT ( 59 * 59 + 50 * 50) = 77.34 m / sec
2. What maximum height was attained by the ball?
Distance_y = V_y * T - 0.5 * 9.8 * T^2
To shoot cannonball up and reach the ground, set Distance_y = 0
0 = V_y * T - 4.9 * T^2
0 = 59 * T - 4.9 * T^2 => 59 * T = 4.9 * T^2 => 59 = 4.9 * T ==> T = 59 / 4.9 ==> T = 12.05 seconds, so it should reach maximum height at 6.025 seconds
Vertical Distance at 6.025 sec,
distance_y = 59 * 6.025 - 0.5 * 9.8 * 6.025^2 = 177.6 m
3. What is the direction of the ball’s velocity just before it strikes the given point?
Vx = 50 m / s
Vyf^2 = Vyi^2 - 2 * 9.8 * Distance_y
Vyf^2 = 59 * 59 - 19.6 * 100
Vyf^2 = 3481 - 1960 = 1519 ==> Vyf = sqrt(1519) = 39 m / s
Direction = ArcTan ( 39 / 50) = 37.95 degree above the horizon
