Can anyone help with the steps of the solutions? A certain airplane can accelera

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  Can anyone help with the steps of the solutions?

A certain airplane can accelerate on the runway at a rate of 2.5 m/s and must attain a speed of 30 m/s in order to take off. What is the minimum length of a runway on which this plane could take off? [Answer: 180.0m ] A ball is thrown straight down from the top of a cliff with a speed of 5 m/s. The cliff-top is 20 m above ground level. Neglect the effects of air resistance. How long is the ball in the air? With what speed does the ball strike the ground? [Answers: (a) 1.57 s (b) 20.42 m/s] A ball is thrown from atop a cliff with an initial speed of 30 m/s at an angle of 40 degree above the horizontal. The cliff is 40 m above level ground. Neglect the effects of air resistance. How long is the ball in the air? Where does the ball strike the level ground? What is the speed and direction of the ball's velocity vector just before it strikes the ground [Answers: (a) 5.44 s (b) 125 m from the base of cliff (c) 41.0 m/s, -55.9 degree] Consider the two vectors A and B Ax=6, Av=5. Bx=-7, By= -1. Find the magnitude and direction of A + B [Answer: 4.12, 104degree ] The position of an object as a function of time is given by x = At1 - Bt + C, where A = 8.0 m/s2, B = 6.0 m/s, and C = 4.0 m. Find the instantaneous velocity and acceleration at t=2s. [Answers: 26 m/s, 16.0 m/s2] What are the period and speed of the motion of a person on a carousel if the person has an acceleration magnitude of 0,80 m/s2 when she is standing 4.0 m from the axis? (b) What are her acceleration magnitude and speed if she then moves in to a distance of 2.0 m from the carousel center and the carousel keeps rotating with the same period? [Answers: (a) 14s, 1.8 m/s, (b) 0.89 m/s, 0.40m/s2]

Explanation / Answer

3) Acceleration = 2.5 m/s^2 and Final Velocity v = 30 m/s

So min length of runway s = v^2 - u^2 / 2a

= 30 ^2 / 5

= 180 m

4)

a) Intial velocity u = 5 m/s and s = 20m

So from s= ut + 0.5* a*t^2 we get 20 = 5t + 4.9 t^2

On solving we get t = 1.573 , -2.593

Since t cant be negative Time it takes to reach ground =1.573 sec

b) Final Speed = v^2 = u^2 + 2 * g * s

so v^2 = 25 + 392

So v= 20.42 m/sec

5) Here therere are basically 2 projectile motions

A ) Time the ball is in air = t1 + t2 = 2 *v* sin(40) /g + t2 = 3.935 + t2

From s = v t + 0.5*g* t^2 we get 40 = v t + 4.9t^2

u is the velocity once it reaches the same height towards the ground = 30 sin40 = 19.28 m/sec

From which we get t2 = 1.5sec

So total time = 3.935 + 1.5 = 5.435 sec

B) Distance where it hits the ground = R1 +R2

= v^2 sin(80) / g + R2

= 90.44 + R2

R2 = ut

So R2 = v cos(40 ) * 1.5

R2 = 34.47 m/sec

So total Range = 90.44 + 34.47 = 124.911 m from the base of cliff

C) The horizontal component of velocity will be same during the entire process which is vcos(40) = 22.981 m/sec

So Vx = 22.981 m/s

Vertical component Vy = Uy +g*t = v sin(40) + 9.8* 1.5

= 19.28 +14.7

= - 33.98 m/sec (Since the direction is opposite)

So spped before it hits ground = sqrt( Vx^2 + Vy^2 )

= 41.02 m/sec

And angle = tan-1(Vy / Vx) = - 55.929 degree

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