Question

A cannon tilted at 25" above the horizontal is fired from the top of a cliff 100 m exits the cannon at a speed of 150 m/s. The ground below is flat. tall. The ball Wh at is the highest point above the ground the ball will reach? What is the speed of the ball at the highest point? How long does it take the ball to reach the ground? How far horizontally will the ball have traveled when it reaches the ground? What is the speed of the ball as it hits the ground? What is the direction of the ball's velocity when it hits the ground? a. b. c. d. e. f.

Answer 1

Given $\theta =25^{\circ}$

Initial velocity, $u=150m/s$

Initial horizontal velocity,$u_{x}=ucos\theta =150*cos25$

$u_{x}=136m/s$

Initial vertical velocity,$u_{y}=usin\theta =150*sin25$

$u_{y}=63.4m/s$

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Consider the vertical motion ,Rising Part

Final vertical velocity =0m/s

Use formula $v^{2}-u^{2}=2as$

$v_{y}^{2}-u_{y}^{2}=-2gy$

$(0m/s)^{2}-(63.4m/s)^{2}=-2*9.81m/s^{2}*y$

$-63.4^{2}=-2*9.81*y$

$y=204.87m$

From Ground ,height $H=100m+y=100m+204.87m =304.87m$

a. ANSWER: ${\color{Red} H=304.87m}$

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at Highest point $v_{y}=0m/s$ and $v_{x}=u_{x}=136m/s$ (there is no acceleration in horizontal direction)

$v_{H}=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{0^{2}+v_{y}^{2}}=v_{y}$

b.ANSWER:${\color{Red} v_{H}=136m/s}$

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c.

Consider vertical motion,Rising Part

Use Formula $v=u+at$

$v_{y}=u_{y}-gt_{1}$

$0m/s=63.4m/s-9.81m/s^{2}*t_{1}$

${\color{Red} t_{1}=6.46s}$

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Falling Part

Use Formula $s=ut+1/2at^{2}$

$H=0m/s*t_{2}+1/2*g*t_{2}^{2}$

$304.87m=0+0.5*9.81*t_{2}^{2}$

$304.87m=4.905t_{2}^{2}$

${\color{Red} t_{2}=7.88s}$

$t=t_{1}+t_{2}=6.46s+7.88s$

c.ANSWER: ${\color{Red} t=14.34s}$

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d.

Speed =Distance/Time

Distance=Speed*Time

Since there is no acceleration in horizontal direction ,horizontal velocity remains same

$X=u_{x}*t$

$X=136m/s*14.34s$

d.ANSWER: ${\color{Red} X=1950.24m}$

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e.

Consider vertical motion,Falling Part

Use Formula $v^{2}-u^{2}=2as$

$v_{y}^{2}-u_{y}^{2}=2gH$

$v_{y}^{2}-(0m/s)^{2}=2*9.81m/s^{2}*304.87m$

$v_{y}=\sqrt{2*9.81*304.87}$

${\color{Red} v_{y}=77.34m/s}$(downwards)

${\color{Red} v_{x}=136m/s}$

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(136m/s)^{2}+(77.34m/s)^{2}}$

e.ANSWER: ${\color{Red} v=156.45m/s}$

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f.

$\theta =tan^{-1}(v_{y}/v_{x})$

$\theta =tan^{-1}(77.34/136)$

f.ANSWER: ${\color{Red} \theta =29.63^{\circ}}\ below \ horizontal\ axis$

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