Answer:
speed of the spaceship is 0.495251 c
Time T = 1.4982 hr
Explanation:
given data
length L1 = 1600 m
length L2 = 1390 m
to find out
speed of the spaceship and time
solution
we know here
L2 = L1 ×
so we find V by put L1 and L2
1390 = 1600 ×
V = 0.495251 c
speed of the spaceship is 0.495251 c
and
we know T1 = 1.95 hr
so we find T2
T1 = T2
put here V and T1 so we get T2
1.95 = T2
Time T = 1.4982 hr
Answer:
30.67m
Explanation:
Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;
s = ut ± ------------(i)
Where;
s = horizontal/vertical displacement
u = initial horizontal/vertical component of the velocity
a = acceleration of the projectile
t = time taken for the projectile to reach a certain horizontal or vertical position.
Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;
h = vt ± ------------(ii)
Where;
h = vertical displacement
v = initial vertical component of the velocity = usinθ
a = acceleration due to gravity (since vertical motion is considered)
t = time taken for the projectile to hit the target
<em>From the question;</em>
u = 47m/s, θ = 0.6rads
=> usinθ = 47 sin 0.6
=> usinθ = 47 x 0.5646 = 26.54m/s
t = 1.7s
Take a = -g = -10.0m/s (since motion is upwards against gravity)
Substitute these values into equation (ii) as follows;
h = vt -
h = 26.54(1.7) -
h = 45.118 - 14.45
h = 30.67m
Therefore, the vertical distance is 30.67m