A projectile is launched from the ground at a speed of 40ms^-1 at an angle of 30 degrees to the horizontal, where does it land? What is the highest point the projectile reaches?

Since the SUVAT equations of motion act independently in the horizontal and vertical directions we can use the vertical equations to find the total time (the time when total vertical displacement is zero) and then calculate how far the partical has moved horizontally in that time.

Vertical:

s=ut+1/2at^2 ------> 0 = 40sin(30) t -0.59.8t^2----> t=4.08s

Horizontal:

s=ut+0.5at^2 -------> x=40cos(30)*4.08  [no horizontal acceleration] =141.3m

To find the apex of the flight, we use a second SUVAT equation, noting that the instantaneous velocity at the maximum is entirely horizontal.

Vertically:

v^2=u^2+2aS------>0=(40sin(30))^2-29.8H------------->H=400/19.6=20.4m

BL
Answered by Ben L. Physics tutor

3111 Views

See similar Physics A Level tutors

Related Physics A Level answers

All answers ▸

Draw a diagram of the forces acting on the rocket as it flies vertically upwards, the rocket is flying through air not a vacuum (it's not in space yet!)


What are vectors?


What is a stationary wave?


If a ball is launched at ground level at a velocity v and angle θ, find an expression for it's height at horizontal distance x.


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences