Question

A hiker throws a rock from ground level at 11.3 m/s at an angle of 26 degrees over a river that is 6.27 m wide. What height is the rock at when it reaches the other side of the river?

Answer 1

For this kind of questions, you are required to know the equation of trajectory for a projectile, given as

Y = X*tan - g*X^{​​​​​​2}/{2*U^{​​​​​​2}*(Cos)²}

Where Y is the height of the projectile from the ground,

X is the horizontal displacement of the projectile,

is the angle of projection,

U is the initial speed of the projectile, &

g is the acceleration due to gravity.

Hence in this question,

You are required to calculate Y (as height of the projectile),

And given data,

X = 6.27 m

U = 11.3 m/s

g = 9.8 m/s^{​​​​​​2},

And = 26°

Substituting these values in equation,

H = 6.27*tan(26°) - 9.8*(6.27)²/{2*(11.3)²*(cos(26°))²}

H = 3.06 - 1.87

H = 1.19m