Question

A projectile is fired at an upward angle of 45.0º from the top of a 265-m cliff with a speed of .sm 185 What will be its speed when it strikes the ground below? (Use conservation of energy.)

Answer 1

The final velocity of the projectile when it strikes the ground below is 198.51 m/s.

Time of motion of the projectile

The time taken for the projectile to fall to the ground is calculated as follows;

h = vt + ¹/₂gt²

where;

• h is height of the cliff
• v is velocity
• t is time of motion

265 = (185 x sin45)t + (0.5)(9.8)t²

265 = 130.8t + 4.9t²

4.9t² + 130.8t - 265 = 0

solve the quadratic equation using formula method,

t = 1.89 s

Final velocity of the projectile

vyf = vyi + gt

where;

• vyf is the final vertical velocity
• vyi is initial vertical velocity

vyf = (185 x sin45) + (9.8 x 1.89)

vyf = 149.322 m/s

vxf = vxi

where;

• vxf is the final horizontal velocity
• vxi is the initial horizontal velocity

vxf = 185 x cos(45)

vxf = 130.8 m/s

vf = √(vyf² + vxf²)

where;

• vf is the speed of the projectile when it strikes the ground below

vf = √(149.322²  +  130.8²)

vf = 198.51 m/s

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