Still trying to escape from the Temple of Gloom, Tennessee Jones has taken refug

Question

Still trying to escape from the Temple of Gloom, Tennessee Jones has taken refuge on a ledge that is 10.0 m above the cavern floor one of his pursuers throws a hand grenade straight up at him with an initial velocity of 14.0 m/s. The release point (point B) of the grenade is 2.10 m above the cavern floor point A) At the same instant he grenade is launched upward Tennessee Jones drops a big rock point c, l0.0 m above the cavern floor hoping to deflect the grenade on its flight upward If the rock and grenade collide, how far above the cavern floor will that occur (m) How fast is each object travelling when they collide?(m/s) If the grenade should miss colliding with the rock, high will it go (m) [measured from the cavern floor]

Explanation / Answer

here,

hc = 10 m

hb = 2.1 m

initial speed of grenade , u1 = 14 m/s

a)

let the rock height when they collide be h' and time be t

h' - hb = u * t - 0.5 * g * t^2

h' - 2.1 = 14 * t - 4.9 * t^2 ....(1)

and

(hc - h') = 0 +0.5 * g * t^2

( 10 - h') = 4.9 * t^2 ....(2)

from (1) and (2)

h' = 8.44 m

t = 0.56 s

the rock and grenade collide 8.44 m from the cavern floor

b)

speed of grenade , v1 = u1 - g * t

v1 = 14 - 9.8 * 0.56 = 8.5 m/s

speed of rock , v2 = g * t

v2 = 9.8 * 0.56 = 5.49 m/s

c)

h4 - hb = u^2 /2g

h4 - 2.1 = 14^2 /2g

h4 = 12.1 m

the height reached is 12.1 m

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