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a projectile is fired at at an angle of 53 degrees with the horizontal. the speed of the projectile is 200 m/s. what is the time the shell will remain in the air?

physics
Guest Jun 2, 2015

Best Answer 

 #1
avatar+26329 
+10

Assuming no air resistance and the projextile is fired from ground level, you can use the kinematic equation applied to the vertical component of motion:

 

s = ut + (1/2)at2  where s is the net distance travelled in time t; u is the initial velocity and a is acceleration.

 

Here: s = 0  (because it starts and ends at ground level)

         u = 200*sin(53°) m/s    (the vertical component of the initial velocity)

         a = -9.8 m/s2    (acceleration due to gravity - acting downwards)

 

This results in a quadratic equation for t

0 = ut + (1/2)at2

or

0 = t(u + (1/2)at)

 

so t = 0 (start time) or

t = -2u/a

 

$${\mathtt{t}} = {\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{200}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{53}}^\circ\right)}}{{\mathtt{9.8}}}} \Rightarrow {\mathtt{t}} = {\mathtt{32.597\: \!367\: \!757\: \!020\: \!408\: \!2}}$$

 

t ≈  32.6 seconds

.

Alan  Jun 3, 2015
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1+0 Answers

 #1
avatar+26329 
+10
Best Answer

Assuming no air resistance and the projextile is fired from ground level, you can use the kinematic equation applied to the vertical component of motion:

 

s = ut + (1/2)at2  where s is the net distance travelled in time t; u is the initial velocity and a is acceleration.

 

Here: s = 0  (because it starts and ends at ground level)

         u = 200*sin(53°) m/s    (the vertical component of the initial velocity)

         a = -9.8 m/s2    (acceleration due to gravity - acting downwards)

 

This results in a quadratic equation for t

0 = ut + (1/2)at2

or

0 = t(u + (1/2)at)

 

so t = 0 (start time) or

t = -2u/a

 

$${\mathtt{t}} = {\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{200}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{53}}^\circ\right)}}{{\mathtt{9.8}}}} \Rightarrow {\mathtt{t}} = {\mathtt{32.597\: \!367\: \!757\: \!020\: \!408\: \!2}}$$

 

t ≈  32.6 seconds

.

Alan  Jun 3, 2015

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