A projectile is shot from the ground at an angle of pi/4 and a speed of 15 m/s. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

2 Answers
May 5, 2017

The distance is =11.5m

Explanation:

Resolving in the vertical direction uarr^+

initial velocity is u_y=vsintheta=15*sin(1/4pi)

Acceleration is a=-g

At the maximum height, v=0

We apply the equation of motion

v=u+at

to calculate the time to reach the greatest height

0=15sin(1/4pi)-g*t

t=15/g*sin(1/4pi)

=1.08s

Resolving in the horizontal direction rarr^+

To find the horizontal distance, we apply the equation of motion

s=u_x*t

=15cos(1/4pi)*1.08

=11.5m

May 5, 2017

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Here in our case
theta=pi/4
u=15m/s
:.u_(y_1)=15sin(pi/4)=15/sqrt2
u_x=15cos (pi/4)=15/sqrt2
When the projectile will reach maximum height
Velocity in y direction u_(y_2)=0 we
:. Applying equating of motion along y direction we get
(u_(y_2))^2 = (u_(y_1))^2 +2 (-g)h
From this equation we get h=5.74m
Now time required to reach the highest point
t=u_(y_2)/g
:. distance moved in x ditection in this time
d=u_(x)xxu_(y_2)/g
:.d=11.5m
:. Distance from starting point of the projectile till it reaches the maximum height is given by

l=sqrt (d^2+h^2)
:.l=12.8m