If a pizza is round like a ball it will roll off a cliff.If it’s just flat and round it won’t.
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;

True I hope this helps you out
I hope it's not too late, but here you go
Answer:
The beam of light is moving at the peed of:
km/min
Given:
Distance from the isalnd, d = 3 km
No. of revolutions per minute, n = 4
Solution:
Angular velocity,
(1)
Now, in the right angle in the given fig.:

Now, differentiating both the sides w.r.t t:

Applying chain rule:


Now, using
and y = 1 in the above eqn, we get:

Also, using eqn (1),

