sunayanaa
Last Activity: 10 Years ago
qn 1:
- We can solve almost every projectile question using just the basic kinematics formulas.
Let us take the sign convention for this problem as g (acceleration due to gravity) as -10m/s^2 in the upward direction (+y axis) and +10m/s^2 in the downward direction (-y axis).
Let initial vel be U
U=50m/s (given)
Let time be t
t=4 sec (given)
let x be the angle of projection which we have to find.
Here the 20 m wall is the vertical distance covered by the projectile.
The range of the projectile would be the horizontal distance covered by it but we dont need to worry about it here.
We can split ‘U’ into its rectangular components
the vertical displacement 20 m of the projectile is due to only its vertical component of U (Uy) which is Usinx.
=50sinx
y(vertical displacement)=Uyt+1/2(-g)(t)^2
=Uyt-1/2(g)(t)^2
20=50sinx-1/2(10)(4)^2
= 20=50sinx-(5)(16)
= 20=50sinx-80
= 50sinx=100
= sinx=1/2
= x=30 degrees
Hope this helps!!!!
