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I need help with the following question:

A smooth spherical object is projected horizontally from a point of vertical height H = 25.78 metres above horizontal ground with a launch speed of u = 24.23m/s . Calculate the time of flight of the object, in seconds, from its point of launch till it hits the ground. Assume that air resistance is negligible and take g = 9.81.

Appreciate any comments and feedback on this. Thanks.

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  1. You can ignore the horizontal velocity as it is constant and with no air resistance does not affect the answer
  2. So in effect you can take the object as being dropped with initial vertical velocity $0$
  3. You may have a formula which allows you to find the time at which it travels a given distance with constant acceleration.
  4. If not, then see that the speed downwards at time $t$ is $v=v_0+gt$ where $v_0=0$. You can integrate this to find the distance travelled as a function of time, and then solve to find the time at which the distance is $H$.
  5. You teacher may prefer you to do this the other way round, so you would be looking at negative speeds upwards with $v=v_0-gt$, and start at a distance $H$ and end at a distance $0$.
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Since the initial velocity of the object is horizontal, we can ignore the launch speed. The answer will be same as if we just dropped the object with no initial velocity. So we just solve $$\frac{1}{2}gt^2 = H.$$

This gives $t = \sqrt{2H/g} = \sqrt{2\cdot 25.78 / 9.81} = 5.26\,\text{s}$. Whoops, that's not quite right. Let's try again:

$$t = \sqrt{2H/g} = \sqrt{2\cdot 25.78 / 9.81} = \boxed{2.29\,\text{s}}$$

That's better. (See comments.)

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From your method, shouldn't the answer be 2.29secs? Square-root of 2H/g gives 2.29 – Xavier Mar 19 '12 at 22:52
Oops, so sorry. You are correct. – JohnJamesSmith Mar 20 '12 at 2:21
No problem, thanks for your input. – Xavier Mar 20 '12 at 17:04

An object is fired at 98m/s, 30degrees abouv the horizontal. If the acceleration due to gravity is 9.8m/s^2 downward find the range of projectile and the velocity on impact.

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