A projectile is shot from the edge of a cliff 115 m above ground level with an i

Question

A projectile is shot from the edge of a cliff 115 m above ground level with an initial speed of v0 = 71 m/s at an angle of 35.0? with the horizontal, as shown in the figure:

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

Express your answer to three significant figures and include the appropriate units.

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

Express your answer to three significant figures and include the appropriate units.

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

Express your answers using three significant figures separated by a comma.

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

Express your answer to three significant figures and include the appropriate units.

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

Express your answer to three significant figures and include the appropriate units.

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

Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

part F) indicates that the projectile is shot UP from the cliff edge.

if we take g = 9.8 m/s^2

Vertical component of initial velocity = Vi*sin35 = 40.72 m/s

Horizontal component of Vi = Vi*cos35 = 58.16 m/s

Time taken to reach max height(Ti) ( when Vvertical = 0 )

using equation V = U + g*T

Ti = Vi*sin35 /g = 4.15s

Max Height (H) reached (This is the answer to part F ) found with V^2 = U^2 + (2*g *H)

0 = 40.72^2 - (2*9.8*H)

(minus because the gravity is acting in the opposite direction to the initial upward velocity.)

H = 40.72^2/19.6

H = 84.6 m (answer of F )

Distance dropped = 84.6 + 115 = 199.6 m

Time to fall this distance given by D =U*T + 0.5*g *T^2

D = 0 + 0.5*g*T^2

T^2 = 199.6*2 / 9.8

T = 6.38 s

Total time of flight = time up + time down

= 4.15 + 6.38 = 10.53 s (answer A)

Horizontal distance travelled = V horizontal * time of flight

= 58.16*10.53 = 612.42 m (answer B)

The horizontal component is constant at 58.16 m/s (answer C)

After falling 199.6 m the vertical component is given by

V^2 = 2*g*H (fall started at u = 0 )

V^2 = 2*9.8*199.6

V = 62.54 m/s (this is downwards.)

so V = - 62.54 m/s (answer C)

Magnitude of velocity is given by sqrt(Vdown^2 + Valong^2)

V = sqrt ( 58.16^2 + 62.54^2 ) = 85.4 m/s (answer D)

Angle with the horizontal given by tan(angle) = Vvertical / Vhorizontal

Angle = arctan (62.54 / 58.16 )

Angle = 47.07 degrees (Answer E)

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