Question

A football is kicked at a velocity of 15m/s at an angle of 25 to the horizontal. Calculate: a) total flight time. b) the maximum height. c) the range of the ball

Answer 1

Total time of flight:2usin¢/g if we take g=10m/s²
T=1.2678s
maximum height:u²sin²¢/2g
H=2.009m
range:u²sin2¢/g
R=17.236m

Answer 2

Answer:

$1.29 \ s, 2.05 \ m \ {and} \ 17.58 \ m.$

Explanation:

Initial velocity of object, $v=15 \ m/s.$

Angle from horizontal, $\theta=25^o.$

Now , formula of time of flight, $t = \dfrac{2\times v \times sin\theta}{g}$

Putting all these values in above equation.

$t=\dfrac{2 \times 15 \times sin25^o}{9.8 } \ s=1.29 \ s$

Also, maximum height , $H_{max}= \dfrac{v^2\times sin^2\theta}{2\times g}.$

Putting all required values,

$H_{max}= \dfrac{15^2\times sin^225}{2\times 9.8}=2.05 \ m$

Again, range is given by , $R=\dfrac{v^2\times sin2\theta}{g}$

Putting required values we get,

$R=\dfrac{15^2\times sin(2\times 25)}{9.8}=17.58 \ m$

Hence, this is the required solution.