Answer:
Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s
At the time of collision velocity of ball one is descending.
Explanation:
Velocity of ball 1 = 146 ft/sec = 44.50m/s
The balls are to collide at an altitude of 234 ft
H = 234 ft = 71.32 m
We have equation of motion
v² = u² + 2as
v² = 44.50² + 2 x (-9.81) x 71.32
v = ±24.10 m/s.
Time for each velocity can be calculated using equation of motion
v = u + at
24.10 = 44.50 - 9.81 t , t = 2.07 s
-24.10 = 44.50 - 9.81 t , t = 6.99 s
Since the second ball throws after 2.3 seconds we can avoid case with t = 2.07 s.
So at the time of collision velocity of ball one is descending.
The collision occurs at t = 6.99 s.
Time of flight of ball 2 = 6.99 - 2.3 = 4.69 seconds.
Height traveled by ball 2 = 71.32 m
We need to find velocity
We have
s = ut + 0.5 at²
71.32 = u x 4.69 - 0.5 x 9.81 x 4.69²
u = 38.21 m/s = 125.36 ft/s
Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s
Answer:
v2=139 ft
Explanation:
First we just look at the motion of the first particle. It is moving vertically in a gravitational field so is decelerating with rate g = 9.81 m/s^2 = 32.18 ft/s^2. We can write it's vertical position as a function of time.
h_1=v_1*t-(a*t/2)
We set this equal to 234 ft to find when the body is passing that point, a solve the quadratic equation for t.
t_1,2=v_1±(√v_1^2-4*a/2*h_1)/a=2.57 s, 7.44 s
Since we know the second ball was launched after 2.3 seconds, we know that the time we are looking for is the second one, when the first ball is descending. The second ball will have 2.3 seconds less so the time we further use is t_c = 7.44 - 2.3 = 5.14 s. With this the speed of the second ball needed for collision at given height, can be found.
Solving a similar equation, but this time for v2 to obtain the result.
h_2=234 ft=v2*t_c-(a*t_c^2/2)--->v2=139 ft
