Question #161897

A stone is projected at an angle 60 degrees to the horizontal with the velocity of 30m/s. Calculate the highest point reached, the range, the time taken for flight and the height of the stone at the instant that the path makes an angle 30 degrees with the horizontal


1
Expert's answer
2021-02-10T10:08:28-0500

a) Let's first find the time that projectile takes to reach the highest point:


vy=v0sinθgt,v_y=v_0sin\theta-gt,0=v0sinθgt,0=v_0sin\theta-gt,t=v0sinθg.t=\dfrac{v_0sin\theta}{g}.

The equation for the motion of projectile in the vertical direction can be written as follows:


y=v0sinθ12gt2.y=v_0sin\theta-\dfrac{1}{2}gt^2.

Substituting tt into the equation we can find the highest point reached by the projectile:


ymax=v02sin2θ2g,y_{max}=\dfrac{v_0^2sin^2\theta}{2g},ymax=(30 ms)2sin26029.8 ms2=34.44 m.y_{max}=\dfrac{(30\ \dfrac{m}{s})^2sin^260^{\circ}}{2\cdot9.8\ \dfrac{m}{s^2}}=34.44\ m.

b)-c) Let's first find the time taken for flight:


tflight=2t=2v0sinθg,t_{flight}=2t=\dfrac{2v_0sin\theta}{g},tflight=230 mssin609.8 ms2=5.3 s.t_{flight}=\dfrac{2\cdot30\ \dfrac{m}{s}\cdot sin60^{\circ}}{9.8\ \dfrac{m}{s^2}}=5.3\ s.

Then, we can find the range of the projectile:


x=v0tflightcosθ,x=v_0t_{flight}cos\theta,x=30 ms5.3 scos60=79.5 m.x=30\ \dfrac{m}{s}\cdot5.3\ s\cdot cos60^{\circ}=79.5\ m.

(d) At the point when the path makes an angle of 3030^{\circ} to the horizontal we have:


tan30=vyvx=v02sin2θ2ghv0cosθ,tan30^{\circ}=\dfrac{v_y}{v_x}=\dfrac{v_0^2sin^2\theta-2gh}{v_0cos\theta},h=v02sin2θv0cosθtan302g,h=\dfrac{v_0^2sin^2\theta-v_0cos\theta tan30^{\circ}}{2g},h=(30 ms)2sin26030 mscos60tan3029.8 ms2=34 m.h=\dfrac{(30\ \dfrac{m}{s})^2sin^260^{\circ}-30\ \dfrac{m}{s}\cdot cos60^{\circ}\cdot tan30^{\circ}}{2\cdot9.8\ \dfrac{m}{s^2}}=34\ m.

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