A projectile Is shot from the edge of a cliff h the figure below 245 m above gro

Question

A projectile Is shot from the edge of a cliff h the figure below 245 m above ground level with an initial speed of vo - 135 m/s at an angle of 37.0° with the horlzontal, as shown in 37.0P (a) Determine the time taken by the projectile to hit point P at ground level (b) Determine the range X of the projectile as measured from the base of the cliff (c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity. (Take up and to the right as positive km directions.) horizontal vertical (d) What is the the magnitude of the velocity? m/s m/s 23

Explanation / Answer

(a)

From the kinematic equation,

y = yo + vy*t + (1/2)gt^2

0 = 245 + 135*sin37*t + (1/2)*-9.8*t^2

4.9t^2 - 81.24t - 245 = 0

By solving the quadratic equation,

t = 19.18 s

Time taken by projectile to hit the point P = 19.18 s

(b)

Range, x = vox*t

x = 135*cos37*19.18

x = 2068.6 m

(c)

Horizontal component of velocity,

vx = 135*cos37 = 107.81 m/s

Vertical component,

vy = 135*sin37 - 9.8*19.18

vy = -106.7 m/s

(d)

Magnitude of velocity,

v = sqrt (vx^2 + vy^2)

v = 151.7 m/s

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