A stone was thrown with velocity 20m/s at an angle of 30 degrees from a height h. The stone moves under gravity freely and reaches the floor 5s after thrown. a) Find H, b)the horizontal distance covered

To find the height we first need to find the initial vertical velocity of the stone. This will be equal to 20 * sin(30) = 200.5 = 10m/s. From here we can use SUVAT. In this case s=-h,u=10m/s, v is unknown, a = -9.8m/s^2 (g), t = 5s. So as s = ut + 0.5at^2, so -h = 105 + -4.95^2 = 50 - 122.5 = -72.5. So h = 72.5m. Now we find the horizontal distance. To do this we first need the initial horizontal velocity. This is 20cos(30) . Now we use SUVAT again. s = ut + 1/2 at^2, as the acceleration in a horizontal direction is 0, we can just use s = u*t, so s = 20cos(30) * 5 = 100cos(30) = 86.6m.

Answered by Dominic H. Maths tutor

4738 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Integrate ln(x) wrt dx


Find all the stationary points of the curve: y = (2/3)x^3 – (1/2)x^2 – 3x + 7/6 and determine their classifications.


Calculate the first derivative of f( x)= 3x^3+2x^2-5


Express 6cos(2x) + sin(x) in terms of sin(x), hence solve the equation 6cos(2x) + sin(x) = 0 for 0<x<360


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences