Question

a projectile is thrown with velocity v metre per second making an angle 30 degree with the horizontal
fine:

a) max height reached
b)time of flight
c) the range of the projectile​

Answer 1

Answer:

A projectile is thrown with velocity v at an angle θ  with horizontal.

then, v

x

​

=vcosθ,v

y

​

=vcosθ

we have to find vertical component of velocity of the projectile at height equal to half of the maximum height.

we know, H

max

​

=

2g

u

2

sin

2

θ

​

here u is intial velocity of projectile. but here given initial velocity is v

so, H

max

​

=

2g

v

2

sin

2

θ

​

now, time taken to reach half of maximum height, t

Y=v

y

​

t+

2

1

​

a

y

​

t

2

2g

v

2

sin

2

θ

​

=vsinθ.t−5t

2

200t

2

−40vsinθ.t+v

2

sin

2

θ=0

t=

10

2

​

(

2

​

±1)vsinθ

​

v

y

​

 

′

=v

y

​

+a

y

​

t

    =vsinθ−

2

​

(

2

​

−1)vsinθ

    = 2 vsinθ

​

Explanation: