Question

1. An object at an angle such that the horizontal range is 4 times of the maximum height .what is the angle projection of the object?

2. Football player hits the ball with speed 20 m per second with an angle 30 degree with respect too original direction .the goal post is a distance of 40 m from him. find out whether the ball reach the goal post.

3. If an object is thrown horizontally with an initial speed 10 M per second from the top of the building of height hundred metre. what is horizontal distance covered by the particle?

4. An object is thrown with initial speed 5 M per second set an angle of projection 30 .what is the height and range reached by the particle?

Answer 1

this is not good
ask one question at a time
now i will answer 2 question for your sake

Final answer : Angle of Projection is 45°
STEPS:
1) According to the question,
Horizontal Range = 4 * (Maximum Height) ---(1)

2) We know that when prijectile is fired at angle ( a) ,then
H(max) = u^2( sin^2(a)) / 2g
Horizontal Range, R = u^2 sin(2a)/g
where u is initial velocity.

Using equation (1),
u^2 sin(2a)/g = 4 * u^2sin^2a/2g

===> sin(2a)=2sin^2 (a)
===>2sina cos a =2sin^2 a
===>sina =cos a
====>a=15
hence a is not equal to 0
hence projection angle is 45 degree

3) This is the case of horizontal projectile.
Given, initial velocity , u = 20 m/s
height , h = 100m
use the formula,
Horizontal range , X = u√(2h/g)
So, X = 20 × √{2 × 100/10}
X = 20 × √20 m

Hence, horizontal distance covered by object is 20√20 m