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Lubov Fominskaja [6]
2 years ago
9

A diver jumps up off a pier at an angle of 25° with an initial velocity of 3.2 m/s. How far from the pier will the diver hit the

water? (Assume the level of the water us the same as the pier)
Physics
2 answers:
rjkz [21]2 years ago
6 0

Answer:

0.8 meters.

I just answered this ame question myself on a test I was taking.

IrinaVladis [17]2 years ago
3 0

Answer:

0.80m

Explanation:

The problem requires that we calculate the range covered by the diver, here the diver is assumed to be a particle undergoing projectile motion.

The range, R is given by equation (1);

R=\frac{u^2sin2\theta}{g}......................(1)

where u is the initial velocity, \theta is the angle of projection and g is acceleration due to gravity which is taken as 9.8m/s^2.

Given;

u = 3.2m/s and \theta=25^o.

Substituting these into equation (1), we obtain the following;

R=\frac{3.2^2sin2(25)}{9.8}\\R=\frac{10.24*sin50}{9.8}\\R=\frac{10.24*0.7660}{9.8}\\R=0.80m

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A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.070
likoan [24]

Given Information:

length of slender rod = L = 90 cm = 0.90 m

mass of slender rod = m = 0.120 kg

mass of sphere welded to one end = m₁ = 0.0200 kg

mass of sphere welded to another end = m₂ = 0.0700 kg (typing error in the question it must be 0.0500 kg as given at the end of the question)

Required Information:

Linear speed of the 0.0500 kg sphere = v = ?

Answer:

Linear speed of the 0.0500 kg sphere = 1.55 m/s

Explanation:

The velocity of the sphere can by calculated using

ΔKE = ½Iω²

Where I is the moment of inertia of the whole setup ω is the speed and ΔKE is the change in kinetic energy

The moment of inertia of a rigid rod about center is given by

I = (1/12)mL²

The moment of inertia due to m₁ and m₂ is

I = (m₁+m₂)(L/2)²

L/2 means that the spheres are welded at both ends of slender rod whose length is L.

The overall moment of inertia becomes

I = (1/12)mL² + (m₁+m₂)(L/2)²

I = (1/12)0.120*(0.90)² + (0.0200+0.0500)(0.90/2)²

I = 0.0081 + 0.01417

I = 0.02227 kg.m²

The change in the potential energy is given by

ΔPE = m₁gh₁ + m₂gh₂

Where h₁ and h₂ are half of the length of slender rod

L/2 = 0.90/2 = 0.45 m

ΔPE = 0.0200*9.8*0.45 + 0.0500*9.8*-0.45

The negative sign is due to the fact that that m₂ is heavy and it would fall and the other sphere m₁ is lighter and it would will rise.

ΔPE = -0.1323 J

This potential energy is then converted into kinetic energy therefore,

ΔKE = ½Iω²

0.1323 = ½(0.02227)ω²

ω² = (2*0.1323)/0.02227

ω = √(2*0.1323)/0.02227

ω = 3.45 rad/s

The linear speed is

v = (L/2)ω

v = (0.90/2)*3.45

v = 1.55 m/s

Therefore, the linear speed of the 0.0500 kg sphere as its passes through its lowest point is 1.55 m/s.

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Answer:

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