Question

I need answers for the problems listed below

Answer 1

Answer:

(27)

• Velocity of salmon is 18m/s.

(28)

• Rocket's totle time in the air is 4.48 second.
• Rocket's maximum height is 33.58 meter.
• Rocket will land 91.6 meter away.

(29)

• Final velocity is 23.96m/s.

(30)

• (a) Peak height is 24.8 meter.

• (b) Initial angle is 23.45°.

• (c) Initial velocity is 55.4m/s

Explanation:

Projectile Motion: When a body is launched in air at an angle $\theta$ with the horizontal  with speed $v$ ,It makes a certain trajectory in influence of gravity.

Time of flight of projectile, $T=\frac{2v\sin\theta}{g}$

Maximum height of projectile, $H_m=\frac{v^2\sin^2\theta}{2g}$

Horizontal range of projectile, $R=\frac{v^2\sin2\theta}{g}$

• (27)

Given,

Distance of final position of salmon from base of cliff, $D=90meter$

Height of cliff, $H=122.5meter$

If the salmon is launched horizontally, the vertical component of velocity $v_y$ will be zero.

Using second equation of motion,

$H=v_yt-\frac{1}{2}gt^2 \\122.5=0-\frac{gt^2}{2} \\t=5second$

Let horizontal component of velocity is $v_x$. Time taken to reach 90 meter away from base is 5 second.

So, $v_x=\frac{D}{t}=\frac{90}{5}\\  =18second$

Hence, Velocity, $v=v_x+v_y=0+18\\=18m/s$

• (28)

Given,

Initial velocity, $v=30m/s$

Angle, $\theta=47^0$

(a)

Totle time, $T=\frac{2v\sin\theta}{g}\\=\frac{2\times30\times\sin47^0}{9.8}\\=4.48second$

So, Rocket's totle time in the air is 4.48 second.

(b)

Maximum height, $H_m=\frac{v^2\sin^2\theta}{2g}\\=\frac{30^2\times\sin^247}{2\times9.8}\\33.58meter$

So, Rocket's maximum height is 33.58 meter.

(c)

Range, $R=\frac{v^2\sin2\theta}{g}\\=\frac{30^2\sin(2\times47)}{9.8}\\ 91.6meter$

So, Rocket will land 91.6 meter away.

• (29)

Given,

Since, It started from rest, then initial velocity $u=0$

Acceleration, $a=7m/s^2$

Distance traveled, $S=41meter$

Using third equation of motion

$v^2-u^2=2as\\v=\sqrt{2as}\\=\sqrt{2\times7\times41}\\ =23.96m/s$

So, Final velocity is 23.96m/s.

• (30)

Given,

Range of trebuchet, $R=225meter$

Time of flight, $T=4.5second$

Now,

$T=\frac{2v\sin\theta}{g} =4.5\\v\sin\theta=\frac{4.5\times9.8}{2} =22.05\\H_m=\frac{(v\sin\theta)^2}{2g} =\frac{22.5^2}{2\times9.8}=24.8meter\\\frac{R}{H_m}=\frac{v^2\times\sin2\theta\times2\times9.8}{v^2\times\sin^2\theta\times9.8} =\frac{225}{24.8} \\\cot\theta=\frac{225}{24.8\times4} \\\theta=23.45^0$

Put value $\theta$

$\frac{2v\sin\theta}{g} =4.5\\v=\frac{4.5\times9.8}{2\times\sin23.45}=55.4m/s$

Hence

(a) Peak height is 24.8 meter.

(b) Initial angle is 23.45°.

(c) Initial velocity is 55.4m/s