Answer:
(a) The range of the projectile is 31,813.18 m
(b) The maximum height of the projectile is 4,591.84 m
(c) The speed with which the projectile hits the ground is 670.82 m/s.
Explanation:
Given;
initial speed of the projectile, u = 600 m/s
angle of projection, θ = 30⁰
acceleration due to gravity, g = 9.8 m/s²
(a) The range of the projectile in meters;
(b) The maximum height of the projectile in meters;
(c) The speed with which the projectile hits the ground is;
Answer:
The rock's final speed at the required altitude will be 42.24 m/s.
Explanation:
Let's start by finding the initial vertical speed.
Vertical Speed = 1.61 * Sin (53.2°)
Vertical Speed = 0.8 m/s
We want to know the speed of the rock when it is at an altitude of 91 km.
The total displacement of the rock from its starting position will thus be equal to -91 km
We can use this in the following equation:
t = 4.3918 seconds
Thus it takes 4.3918 seconds to reach the required altitude. We can now find the speed as follows:
Thus the rock's final speed at the required altitude will be 42.24 m/s.
Answer:
hello the diagram related to this question is missing attached below is the missing diagram
Answer :
The magnitude of the electric field = 4KQ / L^2
direction = 45° east to south
Explanation:
The magnitude of the electric field = 4KQ / L^2
direction = 45° east to south
