Answer:
U = 9.1 m/s
Explanation:
from the question we are given the following
time (t) = 1.8 s
angle = 23 degrees
acceleration due to gravity (g) = 9.8 m/s^{2}
let us first calculate the initial velocity (u) which too the first ball to its maximum height from the equation below
v = u + 0.5at
- The final velocity (v) is zero since the ball comes to rest
- The time (t) it takes to get to the maximum height would be half the time it is in the air, t = 0.5 x 1.8 = 0.9
therefore
0 = u - (0.5 x 9.8 x 0.9)
u = 7.9 m/s
for the second ball to get to the maximum height of the first ball, the vertical component of its initial velocity (U) must be the same as the initial velocity of the first ball. therefore
U sin 60 = 7.9
U = 7.9 ÷ sin 60
U = 9.1 m/s
This should help look at the pictures?
Nine times more (squared speed)
Answer: Ax=(Vx-Vox)/(T)
Vx=Vox+Ax*T
Solving for Ax in terms of Vx, Vox, T
Vx-Vox=Ax*t
Ax=(Vx-Vox)/(T)
This is saying the acceleration in the x-direction can be found by taking the difference between the finial and initial Velocity in x-direction and dividing it by the Total Time.
Any questions please feel free to ask. Thanks
