# hey! can someone please help me solve these 2 questions.. i m stuck.. thanks a lot in advance?

1) a 50 g rock is thrown with a velocity of 15 m/s [50 deg from the horizontal] fro the top of a 10 m tower.

a) calc. the speed when the rock is at the max. height.

b) calc. the max. height.

c) calc the velocity when the rock reaches the ground.

2) a 3 kg ball is dropped from the eight of 0.80 m above the top of a vertical spring of force constant 1200 N/m. what is the max compression of the spring?

### 1 Answer

- JimLv 76 years agoFavorite Answer
1) Vo = 15 m/s at Θ = 50°

Voy = 15(sin 50) = 11.5 m/s

Vox = 15(cos 50) = 9.64 m/s

time to reach max height = Voy/g = t = 11.5/9.81 = 1.17227

max height above tower = 1/2gt² = (0.5)(9.81)(1.17227)² ≈ 6.74 m

max height of rock = H = {above ground} = 16.74 m ANS (b)

at max height the rock has only Vox speed = 9.64 m/s ANS (a)

when rock reaches ground:

its vertical component of velocity = √2gH = √(2)(9.81)(16.74) = 4.26 m/s

its horizontal componet of velocity = Vox = 9.64 m/s

its SIZE of final velocity = √(4.26²+9.64²) = 10.5 m/s ANS (c)

its Θ = arctan -4.26/9.64 = -23.8° ANS (c)

2) In such problems when spring is VERTICAL be aware of the GPE loss as spring is compressed

let x = max compression (distance) of spring

SPE = GPE of dropped height + spring compression distance

1/2kx² = mg(h+x) where k = 1200 N/m and h = 0.80 m

600x² = 3(9.81)(0.80 + x) = 23.544 + 29.43x

0 = -600x² + 29.43x + 23.544

solve quadratic for positive x:

x ≈ 0.244 m ANS

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