Answer:
1. v = 22.2 m/s
2. t = 2.25 seconds
3. h = 27.05 m
4. t = 1.16 seconds
Explanation:
The questions involve motion under the influence of gravity
1. Using the formula v² = u² + 2gh
where u = 1.6 m/s; g= 9.81 m/s²; h = 25 m; v = ?
v² = (1.6)² + 2 * 9.81 * 25
√v² = √493.06
v = 22.2 m/s
2. Using h = ut + 1/2 gt²
where h = 25 m; u = 0 (since velocity on reaching the ground is zero); g = 9.81 m/s²; t = ?
therefore, h = 1/2 gt²
making t subject of the formula, t = √ (2*h /g)
t = √ (2 * 25 / 9.81)
t = 2.25 seconds
3. Time of travel for the blue ball, t = 2 - 0.4 = 1.6s
using h = ut - gt²
u = 24 m/s; t = 1.6 s; g = 9.81 m/s²
note: since the ball is travelling against gravity, g is negative
h = 24 * 1.6 - 11/2 * 9.81 * 1.6²
h = 38.4 - 12.55 = 25.85 m
since height above the ground is 1.2 m,
total height h = 25.85 m + 1.2 m
h = 27.05 m
4. Let the time of travel of the red ball be t seconds.
So the time of travel of the blue ball = (t - 0.4) seconds.
Both the balls are at the same height :
25 - s = 1.2 + h where s & h are the displacements of the red & the blue ball respectively.
25 - (ut + 1/2 gt2) = 1.2 + (ut - 1/2 gt2)
25 - (1.6 t + 0.5 * 9.8 t²) = 1.2 + (24(t-0.4) - 0.5*9.8*(t-0.4)²)
solving the equation above for the time after which both the balls are at the same height.
25 - 1.6t - 4.9t² = 1.2 + 24t - 9.6 - 4.9t² + 3.92t - 0.784
collecting like terms
(25 - 1.2 + 9.6 + 0.784) = (24 + 3.92 + 1.6) * t
t = 34.184 / 33.44
t = 1.16 seconds
