a soccer ball is kicked from the ground at an initial velocity of 19.5m/s at an upward angle of 45 degrees. A player 55meters away in the direction of the kick starts running to meet the ball at that instant. What must be his average sped if he is to meet the ball right as it lands?
+33652 Assuming no resistance from the air (unrealistic in the case of a football!) the time of flight is given by:
0 = 19.5*sin(45°)*t - (1/2)*9.8*t2, taking g = 9.8m/s2 (this comes from considering the vertical motion)
$${\mathtt{t}} = {\frac{{\mathtt{19.5}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{45}}^\circ\right)}{\mathtt{\,\times\,}}{\mathtt{2}}}{{\mathtt{9.8}}}} \Rightarrow {\mathtt{t}} = {\mathtt{2.813\: \!996\: \!374\: \!111\: \!530\: \!6}}$$ seconds
The horizontal distance travelled by the ball is s = 19.5*cos(45°)*t
$${\mathtt{s}} = {\mathtt{19.5}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{45}}^\circ\right)}{\mathtt{\,\times\,}}{\mathtt{2.813\: \!996\: \!374\: \!111\: \!530\: \!6}} \Rightarrow {\mathtt{s}} = {\mathtt{38.801\: \!020\: \!408\: \!212\: \!922\: \!5}}$$ metres
The player must therefore run a distance of (55 - s) metres in a time t seconds, so his average speed must be (55 - s)/t m/s
$${\mathtt{v}} = {\frac{\left({\mathtt{55}}{\mathtt{\,-\,}}{\mathtt{38.801}}\right)}{{\mathtt{2.814}}}} \Rightarrow {\mathtt{v}} = {\mathtt{5.756\: \!574\: \!271\: \!499\: \!644\: \!6}}$$ m/s
or v ≈ 5.76 m/s
.
+33652 Assuming no resistance from the air (unrealistic in the case of a football!) the time of flight is given by:
0 = 19.5*sin(45°)*t - (1/2)*9.8*t2, taking g = 9.8m/s2 (this comes from considering the vertical motion)
$${\mathtt{t}} = {\frac{{\mathtt{19.5}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{45}}^\circ\right)}{\mathtt{\,\times\,}}{\mathtt{2}}}{{\mathtt{9.8}}}} \Rightarrow {\mathtt{t}} = {\mathtt{2.813\: \!996\: \!374\: \!111\: \!530\: \!6}}$$ seconds
The horizontal distance travelled by the ball is s = 19.5*cos(45°)*t
$${\mathtt{s}} = {\mathtt{19.5}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{45}}^\circ\right)}{\mathtt{\,\times\,}}{\mathtt{2.813\: \!996\: \!374\: \!111\: \!530\: \!6}} \Rightarrow {\mathtt{s}} = {\mathtt{38.801\: \!020\: \!408\: \!212\: \!922\: \!5}}$$ metres
The player must therefore run a distance of (55 - s) metres in a time t seconds, so his average speed must be (55 - s)/t m/s
$${\mathtt{v}} = {\frac{\left({\mathtt{55}}{\mathtt{\,-\,}}{\mathtt{38.801}}\right)}{{\mathtt{2.814}}}} \Rightarrow {\mathtt{v}} = {\mathtt{5.756\: \!574\: \!271\: \!499\: \!644\: \!6}}$$ m/s
or v ≈ 5.76 m/s
.
