Question

A projectile takes off with an initial velocity of 10 m/s at an angle of elevation 45°

Answer 1

Answer:

IN the above mentioned question the initital velocity of the projectile u and it's angle with the horizontal axis is give, from the given two days we will be able to any details we are asked to do so, such as distance

t= total time for travel

t=2×v×sinФ÷g

H = maximum height (m)

v0 = initial velocity (m/s)

g = acceleration due to gravity (9.80 m/s2)

θ = angle of the initial velocity from the horizontal plane (radians or degrees)

Answer 2

Answer:

Explanation:

y\quad =\quad xtan\theta -\frac { g{ x }^{ 2 } }{ 2{ u }^{ 2 }{ cos }^{ 2 }\theta } \\ 2\quad =\quad x-\frac { 10\times { x }^{ 2 } }{ 2\times { 10 }^{ 2 }\times \frac { 1 }{ 2 } } =x-\frac { { x }^{ 2 } }{ 10 } \\ 2=x-\frac { { x }^{ 2 } }{ 10 } \\ \Rightarrow x=\quad 5\pm \sqrt { 5 } \\ \Rightarrow d={ x }_{ 1 }-{ x }_{ 2 }=2\sqrt { 5 } =4.47m\\ first\quad hurdle\quad is\quad at\quad { x }_{ 1 }\quad =\quad 5-\sqrt { 5 } =2.75m