(A) Determine the time taken by the projectile to hit point P at ground level. (

Question

(A) Determine the time taken by the projectile to hit point P at ground level.

(B) Determine the distance X of point P from the base of the vertical cliff.

(C) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity.

(D)At the instant just before the projectile hits point P, find the magnitude of the velocity.

(E) At the instant just before the projectile hits point P, find the angle made by the velocity vector with the horizontal.

(F) Find the maximum height above the cliff top reached by the projectile.

PHY Homework 4 Problem 3.59 At the instant just befor Express your answer Constants A projectile is shot from the edge of a cliff 115 m above ground level with an initial speed of vo 46 m/s at an angle of 35.0° with the horizontal, as shown in the figure (Figure 1). = value Submit Request Figure 1 of 1 Part F Find the maximum heig Express your answer 35.0 maxValue Submit Request Provide Feedback

Explanation / Answer

1.

Given that

V0 = 46 m/sec at 35 deg with horizontal

V0x = 46*cos 35 deg = 37.68 m/sec

V0y = 46*sin 35 deg = 26.38 m/sec

ay = -g = -9.8 m/sec^2

ax = 0 m/sec^2

Height of cliff = -115 m

Using 2nd kinematic equation

H = V0y*t + 0.5*ay*t^2

-115 = 26.38*t - 0.5*9.8*t^2

4.9*t^2 - 26.38*t - 115 = 0

Solving quadratic equation

t = [26.38 +/- sqrt (26.38^2 + 4*4.9*115)]/(2*4.9)

taking +ve sign

t = 8.23 sec

Part B.

Range in projectile motion is given by:

R = V0x*t

R = 37.68*8.23

R = 310.1 m

Part C.

In projectile motion horizontal velocity remains constant as there is no acceleration in horizontal direction, So

Vfx = V0x = 37.68 m/sec

Now Using 1st kinematic equation

Vfy = V0y + ay*t

Vfy = 26.38 - 9.81*8.23

Vfy = -54.36 m/sec

Part D.

Magnitude of velocity will be

Vf = sqrt (Vfx^2 + Vfy^2)

Vf = sqrt (37.68^2 + (-54.36)^2)

Vf = 66.14 m/sec

Part E.

Direction = arctan (Vfy/Vfx)

= arctan (54.36/37.68)

= 55.27 deg below the horizontal

Part F.

At max height vertical velocity component will be zero, So

Using 3rd kinematic equation

V1y^2 = V0y^2 + 2*ay*Hmax

Hmax = (V1y^2 - V0y^2)/(2*ay)

Hmax = (0^2 - 26.38^2)/(2*(-9.81))

Hmax = 35 47 m from the top of cliff

Hmax = 115 + 35.47 = 150.47 m from the bottom of cliff

See that Hmax is required from the top of cliff, So Hmax = 35.47 m

Also Check that final answer should be in mentioned significant digits, change your answer in correct significant digits.

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