Answer:
17.7 m/s
Explanation:
Given:
y₀ = 0 m
y = 16 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: v
v² = v₀² + 2a (y − y₀)
v² = (0 m/s)² + 2 (9.8 m/s²) (16 m − 0 m)
v = 17.7 m/s
The ball is moving at a speed of 17.7 m/s when it hits the ground.
Explanation:
t = usin©/g
Where t is the time to reach the maximum height
Time spent in air is T = 2t
Hence, T = 2usin©/g
T = 2 x 20 x sin 65°/ 9.8
T = 3.69s
Answer:
a) v = 54.7m/s
b) v = (58 - 1.66a) m/s
c) t = 69.9 s
d) v = -58.0 m/s
Explanation:
Given;
The height equation of the arrow;
H = 58t - 0.83t^2
(a) Find the velocity of the arrow after two seconds. m/s;
The velocity of the arrow v can be given as dH/dt, the change in height per unit time.
v = dH/dt = 58 - 2(0.83t) ......1
At t = 2 seconds
v = dH/dt = 58 - 2(0.83×2)
v = 54.7m/s
(b) Find the velocity of the arrow when t = a. m/s
Substituting t = a into equation 1
v = 58 - 2(0.83×a)
v = (58 - 1.66a) m/s
(c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s
the time when H = 0
Substituting H = 0, we have;
H = 58t - 0.83t^2 = 0
0.83t^2 = 58t
0.83t = 58
t = 58/0.83
t = 69.9 s
(d) With what velocity will the arrow hit the surface? m/s
from equation 1;
v = dH/dt = 58 - 2(0.83t)
Substituting t = 69.9s
v = 58 - 2(0.83×69.9)
v = -58.0 m/s