Bob has finished climbing a sheer cliff a beach, and wants to figure out how hig

Question

Bob has finished climbing a sheer cliff a beach, and wants to figure out how high climbed. he has to a above on the ground below with a long starts atch, and a friend mph. Bob measuring itcher, and knows that the fastest he can throw the ball is 81.0 tch as he throws the ball (with way to measure the ball's initial trajectory), and watches Bob carefully rises falls, and seconds the is ain level with Bob. see well enough to time when the ball hits the ground. Bob's friend then measures that the ball the ft the base of the How high up if the ball started from exactly 5 ft above edge of the cliff? Number 81.0 mph 0, 910 s 321 ft

Explanation / Answer

given

vo = 81 mph

= 81*5280/(60*60) ft/s (since amile = 5280 ft)

= 118.8 ft/s

g = 9.8 m/s^2

= 9.8*3.28 ft/s^2

= 32.1 ft/s^2

let theta is the angle of projection.

time taken to reach the same height, t = 2*voy/g

==> voy = t*g/2

= 0.91*32.1/2

= 14.6 ft/s

now use, vo^2 = vox^2 + voy^2

==> vox= sqrt(vo^2 - voy^2)

= sqrt(118.2^2 - 14.6^2)

= 117.3 ft/s

total time taken for the ball to hit the ground,

T = x/vox

= 321/117.3

= 2.7365 s

let h is the height of the Bob above ground level.

use,

-(h + 5) = voy*t - (1/2)*g*t^2

-(h + 5) = 15.6*2.7365 - (1/2)*32.1*(2.7365)^2

==> h = 72.5 ft <<<<<<<<<-----------------------------Answer

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