The result is:
H = 181.3m
t = 2.9s
Here’s the breakdown of the explanation:
Provided information:
Initial speed V = 49 m/s
Acceleration g = - 9.8 m/s^2
Starting height h = 58.8 m
The equation that captures this scenario
V^2 = u^2 + 2g(h)
To ascertain the maximum height. What's the height
At the peak, V = 0
0 = 49^2 - 2 × 9.8 H
19.6H = 2401
From here, H = 2401/19.6
H = 122.5 - h
Resulting in H = 122.5 + 58.8
Therefore, H = 181.3 m
How many seconds would it take for the arrow to reach the ground?
h = ut + 1/2gt^2
181.3 = 49t + 0.5 × 9.8t^2
181.3 = 49t + 4.9t^2
Utilizing the quadratic formula
t = (-49 - 77.2)/ 9.8 or (-49+ 77.2)/9.8
t = positive
t = 2.9 s
To determine the time when the arrow is approximately 100 up from the cliff’s base using the earlier equation.
.V^2 = u^2 + 2g*h.
V^2 = 0 + 19.6(181.3-100) = 1593.48, V = 39.9 m/s.
To evaluate the height of the arrow after 2 seconds
h = ut + 0.5gt^2.
U = 49 m/s, t = 2 s., g = -9.8 m/s^2, h h = 98 + 19.6
h = 78.4
