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?
+33615 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
.
+33615 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
.
