A projectile is fired horizontally from the edge of a cliff, above the ocean, with an initial velocity of 110 m/s.The height of the cliff is 115 metres.Take gravitational acceleration to be 9.81 m/s2.Hint 1: Draw a diagram to identify known and unknown valuesHint 2: Set your datum horizontal at water levelHint 3: Neglect the effect of air resistanceHint 4: Remember sign convention when dealing with vectorsDetermine the following:(a) how many seconds it will take for the projectile to hit the water, t = Answer seconds,(b) the horizontal distance from the base of the cliff will the projectile hit the water, sx= Answer metres,(c) the final vertical velocity component of the projectile the moment it hits the water, vy = Answer m/s.
Question
A projectile is fired horizontally from the edge of a cliff, above the ocean, with an initial velocity of 110 m/s.The height of the cliff is 115 metres.Take gravitational acceleration to be 9.81 m/s2.Hint 1: Draw a diagram to identify known and unknown valuesHint 2: Set your datum horizontal at water levelHint 3: Neglect the effect of air resistanceHint 4: Remember sign convention when dealing with vectorsDetermine the following:(a) how many seconds it will take for the projectile to hit the water, t = Answer seconds,(b) the horizontal distance from the base of the cliff will the projectile hit the water, sx= Answer metres,(c) the final vertical velocity component of the projectile the moment it hits the water, vy = Answer m/s.
Solution
(a) To find the time it will take for the projectile to hit the water, we can use the equation of motion: h = 0.5gt^2, where h is the height of the cliff and g is the acceleration due to gravity. Rearranging for t gives us t = sqrt(2h/g). Substituting the given values gives us t = sqrt((2*115)/9.81) = 4.83 seconds.
(b) The horizontal distance the projectile will travel before hitting the water can be found using the equation s = ut, where u is the initial horizontal velocity and t is the time. Substituting the given values gives us s = 110*4.83 = 531.3 metres.
(c) The final vertical velocity of the projectile can be found using the equation v = gt, where g is the acceleration due to gravity and t is the time. Substituting the given values gives us v = 9.81*4.83 = 47.4 m/s.
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