Question

A projectile is fired from ground level with a velocity of 500 m/s at 30° to the horizontal. Calculate its horizontal range, the greatest height it reaches, and the time taken to rise to that height.

Answer 1

Answer on Question 49948, Physics, Mechanics — Kinematics — Dynamics A projectile is fired from ground level with a velocity of 500 m/s at 30 to the horizontal. Calculate its horizontal range, the greatest height it reaches, and the time taken to rise to that height.

Solution

Vertical initial speed is

$v_{y0}=v_{0}\sin\alpha=500\cdot\sin 30{}^{\circ}=250\,m/s$

Time it takes particle to reach highest point can be found from equation for velocity

$v_{y}=v_{y0}-gt$

At that point velocity is zero, hence

$t_{h}=\frac{v_{y0}}{g}=\frac{500}{9.8}\approx 25.5\,s$

The height itself is equal to

$h=\frac{v_{y0}^{2}}{2g}\approx 3188.9\,m$

Now we can find horizontal distance. It is equal to twice the time $t_{h}$ multiplied by horizontal velocity

$l=2v_{x0}t_{h}=2v_{0}\cos\alpha\cdot t_{h}=2\cdot 500\cdot\sin 30{}^{\circ}\cdot 25.5\approx 22083.6\,m$

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