Answer:
2.1 s
Explanation:
The motion of the ball is a projectile motion. We know that the horizontal range of the ball is
And that the initial speed of the ball is
at an angle of
So, the horizontal speed of the ball (which is constant during the entire motion) is
And since the horizontal range is 50 m, the time taken for the ball to cover this distance was
which is the time the ball spent in air.
Answer:
The second ball lands 1.5 s after the first ball.
Explanation:
Given;
initial velocity of the ball, u = 12 m/s
height of fall, h = 35 m
initial velocity of the second, v = 12 m/s
Time taken for the first ball to land;
determine the maximum height reached by the second ball;
v² = u² -2gh
at maximum height, the final velocity, v = 0
0 = 12² - (2 x 9.8)h
19.6h = 144
h = 144 / 19.6
h = 7.35 m
time to reach this height;
Total height above the ground to be traveled by the second ball is given as;
= 7.35 m + 35m
= 42.35 m
Time taken for the second ball to fall from this height;
total time spent in air by the second ball;
T = t₁ + t₂
T = 1.23 s + 2.94 s
T = 4.17 s
Time taken for the second ball to land after the first ball is given by;
t = 4.17 s - 2.67 s
T = 1.5 s
Therefore, the second ball lands 1.5 s after the first ball.