Physics projectiles help-please?

Question

A projectile was fired across level ground and reached a maximum height of 26.14 metres. What was the time of flight of this projectile?

Give your answer in seconds (s) and round it to three (3) significant figures.

Time of Flight (s) =

A projectile was fired across level ground with an initial velocity of 22.26 metres per second at an angle of 51.00 degrees above the horizontal. What was the maximum height of this projectile?

Give your answer in metres (m) and round it to two (2) decimal places using the normal convention. Pad with zeros if necessary.

Maximum Height (m) =

Answer 1

1) Maximum height = u^2 sin^2(@) / 2g                (where @ is the angle of projection)

So

26.14 = u^2 sin^2(@) / 2g

u^2 sin^2(@) = 26.14 * 2 * 9.8

u sin@ = 22.635

T = 2u sin@ / g = 4.619 s

2) u = 22.26

@ = 51

Maximum height = u^2 sin^2 ( @ ) / 2g = 15.269

Hope this helps.

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