A smooth spherical object (the first object) is projected horizontally from a po

Question

A smooth spherical object (the first object) is projected horizontally from a point a vertical height H = 26.94 metres above horizontal ground with a launch speed u = 21.63 ms-1.
A second identical object is projected from ground level with launch speed v and at an angle ? above the horizontal. Calculate the value of this launch angle ? in radians, if the second object is to have the same horizontal range and the same time of flight as the first object. Please show working and give your answer to 3 decimal places and take g = 9.81 ms-2.

Explanation / Answer

To have both the same time and range both projectiles must have the same horizontal speed.
Vx = 21.63 m/s
H = 1/2 g t^2
t = sqrt(2 H/g) = sqrt(2 * 26.94 / 9.81) = 2.343 sec
range = x = Vx t = 21.63 * 2.343
range = 50.7 m

Vy = 1/2 gt
(The one half is because half the time it is going up and half the time down so the change is gt)
Vy = 1/2 * 9.81* 2.343 s = 11.5 m/s
? = arctan (Vy/Vx)= arctan (11.5 / 21.63 ms-1) = 28o

alternately, (height of dropped object with no Vy component) / (range) gives equivalent launch angle
? = arctan (26.94 / 50.7) = 28o

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