Is fired horizontally with a velocity of 98 metre per second from the top of the hill 490 metre high find the velocity with which projectile strikes the ground?
Answers
Answer:
1.) Time taken to reach the ground:
S = ut + 0.5gt²
U = 98
g = 10
490 = 98t + 5t²
5t² + 98t - 490 = 0
t = ( -98 +/- √9604 + 9800) / 10
t = (-98 -/+ 139.30) / 10
t = 4.13 seconds.
2.) Range = ut
Range = 98 × 4.13 = 404.74 m
3.) Final velocity:
V² = U² + 2gs
V² = 9604 + 9800
V² = 19404
V = 139.30
Answer:
Figure of the projectile motion in the attachment.
The initial velocity of the body = 98 m/s.
The height of the hill = 490 m.
Find the velocity at which the body strikes the ground = ?
Finding the time :
⇒ s_y = (u_y)t + ½(a_y)t²
⇒ 490 = 0 + ½(9.8)(t²)
⇒ t = 10 s
We know that, v_x = u_x + (a_x)t
⇒ v_x = 98 + 0 = 98 m/s
And, v_y = u_y + (a_y)t
⇒ v_y = 0 + (9.8)(10) = 98 m/s
So, v = √[(v_x)² + (v_y)²]
⇒ v = √(98)² + (98)²
⇒ v = 98√2 m/s
Hence, the speed at which the body will strike the ground = 98√2 m/s.
