Answer:
a) 69.3 m/s
b) 18.84 s
Explanation:
Let the initial velocity = u
The vertical and horizontal components of the velocity is given by uᵧ and uₓ respectively
uᵧ = u sin 40° = 0.6428 u
uₓ = u cos 40° = 0.766 u
We're given that the horizontal distance travelled by the projectile rock (Range) = 1 km = 1000 m
The range of a projectile motion is given as
R = uₓt
where t = total time of flight
1000 = 0.766 ut
ut = 1305.5
The vertical distance travelled by the projectile rocks,
y = uᵧ t - (1/2)gt²
y = - 900 m (900 m below the crater's level)
-900 = 0.6428 ut - 4.9t²
Recall, ut = 1305.5
-900 = 0.6428(1305.5) - 4.9 t²
4.9t² = 839.1754 + 900
4.9t² = 1739.1754
t = 18.84 s
Recall again, ut = 1305.5
u = 1305.5/18.84 = 69.3 m/s
Answer:
C.) The Distance DH = 1.5 lambda
Explanation:
This statement C.) is false, because it does not count as the 1.5 wavelength, it is less than 1 wavelength.
Total distance = 76+54 = 130km
total time = 2+5 = 7hrs
Av. speed = 130/7 = 18.571km/hr = 18.6 km/h ( 3 sig fig)
