Math Projectile Question

Question

There is a cannon on a cliff over the ocean. The cliff is 64 ft high. The cannon shoots the cannonball out at 100ft/sec. How far does it reach? 

Answer 1

You can solve this in two parts, first considering the vertical and then calculating the horizontal. First, find out how many seconds it will take for the cannonball to reach the ground. Every t seconds, an object dropped from a fixed position moves vertically by -16t^2 feet.

64 - 16t^2 = 0

64 = 16t^2

4 = t^2

2 = t

The cannonball is in the air for 2 seconds and it travels 100 ft/sec, so the cannonball travels 200 feet.

Answer 2

H(t) = - (1 / 2) g t ² + Vo t + Ho

Ho = 64 ft

Vo = 100 ft/s

g = 32.2 ft/s^2

H(T) = -0.5*32.2[T ^2 ] + 100[T] + 64

= 64-16.1T^2+100T

PLOT the upside parabolic function

OR factor to find when H(T) = 0 thus find two times when height is zero aka hits ocean.

2 roots are T = 6.7961 seconds and -0.584917 seconds

negative T DNE so T = 6.7961 seconds

Distance = Dx = Vx*T= 100ft/sec * 6.7961 sec

Distance = 679.61 feet

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NOTE ANGLE SHOT OUT OF IS IMPORTANT!!!!

assuming angle was horizontal to cliff or angle was 0 degrees

then answer is correct next time denote initial conditions

Vx Vy time at rest, angle ,....

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